Atomic Theory And Structure
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Calculations Involving Moles
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[{"Name":"Atomic Theory And Structure","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Daltons Atomic Theory","Duration":"8m 30s","ChapterTopicVideoID":16882,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16882.jpeg","UploadDate":"2019-02-19T05:40:56.3470000","DurationForVideoObject":"PT8M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.020","Text":"In this video, we will discuss some of the earliest laws of"},{"Start":"00:04.020 ","End":"00:08.070","Text":"chemistry that formed the basis of Dalton\u0027s atomic theory."},{"Start":"00:08.070 ","End":"00:12.660","Text":"The first law is the law of conservation of mass,"},{"Start":"00:12.660 ","End":"00:18.435","Text":"which was formulated by the French chemists Lavoisier in 1774."},{"Start":"00:18.435 ","End":"00:23.580","Text":"This law says that the total mass of substances present after"},{"Start":"00:23.580 ","End":"00:26.130","Text":"a chemical reaction is the same as"},{"Start":"00:26.130 ","End":"00:31.169","Text":"the total mass of substances before a chemical reaction."},{"Start":"00:31.169 ","End":"00:36.615","Text":"Matter is neither created nor destroyed in a chemical reaction."},{"Start":"00:36.615 ","End":"00:41.370","Text":"To illustrate this, let\u0027s use the example of a light bulb."},{"Start":"00:41.370 ","End":"00:47.290","Text":"Consider an old-fashioned incandescent light bulb."},{"Start":"00:50.770 ","End":"00:54.770","Text":"The tungsten metal filament is heated to"},{"Start":"00:54.770 ","End":"00:59.105","Text":"a high temperature and gives off light, here\u0027s the light."},{"Start":"00:59.105 ","End":"01:05.615","Text":"After prolonged use, the filament becomes oxidized and breaks."},{"Start":"01:05.615 ","End":"01:09.185","Text":"It has undergone a chemical reaction."},{"Start":"01:09.185 ","End":"01:13.550","Text":"If we compare the mass of the light bulb when we just"},{"Start":"01:13.550 ","End":"01:17.510","Text":"bought it with that which has been destroyed,"},{"Start":"01:17.510 ","End":"01:19.360","Text":"which is no longer working,"},{"Start":"01:19.360 ","End":"01:22.280","Text":"we\u0027ll discover something very interesting."},{"Start":"01:22.280 ","End":"01:27.370","Text":"We will discover that the mass is identical."},{"Start":"01:37.360 ","End":"01:42.230","Text":"The second law is the law of constant composition,"},{"Start":"01:42.230 ","End":"01:46.745","Text":"which was formulated by Proust in 1799."},{"Start":"01:46.745 ","End":"01:50.520","Text":"This is not the same Proust has wrote the famous novel,"},{"Start":"01:50.520 ","End":"01:52.135","Text":"In Search of Lost Time,"},{"Start":"01:52.135 ","End":"01:53.900","Text":"but a different French."},{"Start":"01:53.900 ","End":"01:56.000","Text":"What does this law say?"},{"Start":"01:56.000 ","End":"02:01.625","Text":"This law says that all samples of a compound have the same composition."},{"Start":"02:01.625 ","End":"02:04.220","Text":"It doesn\u0027t matter where you take them from,"},{"Start":"02:04.220 ","End":"02:06.895","Text":"they will always be the same."},{"Start":"02:06.895 ","End":"02:09.500","Text":"Let\u0027s take an example."},{"Start":"02:09.500 ","End":"02:13.280","Text":"The example I\u0027ve chosen is that of ascorbic acid,"},{"Start":"02:13.280 ","End":"02:15.725","Text":"often called vitamin C,"},{"Start":"02:15.725 ","End":"02:21.110","Text":"which has the formula C_6H_8O_6."},{"Start":"02:21.110 ","End":"02:24.080","Text":"That means that in every molecule,"},{"Start":"02:24.080 ","End":"02:26.195","Text":"there are 6 carbons,"},{"Start":"02:26.195 ","End":"02:31.505","Text":"8 hydrogens, and 6 oxygens."},{"Start":"02:31.505 ","End":"02:34.550","Text":"In this picture, we see an orange,"},{"Start":"02:34.550 ","End":"02:37.385","Text":"which is an excellent source of vitamin C,"},{"Start":"02:37.385 ","End":"02:42.805","Text":"and pills which people often take when they have a bad cold."},{"Start":"02:42.805 ","End":"02:50.270","Text":"Now, if we were to examine the vitamin C in either the orange or in the pills,"},{"Start":"02:50.270 ","End":"02:53.165","Text":"we\u0027d find something very interesting."},{"Start":"02:53.165 ","End":"02:59.750","Text":"We would find that each of them has precisely the same structure,"},{"Start":"02:59.750 ","End":"03:06.295","Text":"precisely the same composition as well: 41 percent carbon,"},{"Start":"03:06.295 ","End":"03:09.465","Text":"4.5 percent hydrogen,"},{"Start":"03:09.465 ","End":"03:13.165","Text":"and 54.5 percent oxygen."},{"Start":"03:13.165 ","End":"03:21.830","Text":"It doesn\u0027t matter whether you get your vitamin C out of oranges or out of pills,"},{"Start":"03:21.830 ","End":"03:29.020","Text":"they both have precisely the same structure and precisely the same composition."},{"Start":"03:29.020 ","End":"03:36.380","Text":"Now, Dalton took these 2 laws and wrote his Dalton\u0027s atomic theory."},{"Start":"03:36.380 ","End":"03:43.130","Text":"He did this in the early 19th century in 1803-1808."},{"Start":"03:43.130 ","End":"03:47.480","Text":"Now, Dalton\u0027s atomic theory has 3 parts."},{"Start":"03:47.480 ","End":"03:49.130","Text":"The first part says,"},{"Start":"03:49.130 ","End":"03:51.920","Text":"\"Each chemical element is composed of"},{"Start":"03:51.920 ","End":"03:56.510","Text":"atoms that cannot be created or destroyed in a chemical reaction.\""},{"Start":"03:56.510 ","End":"04:02.160","Text":"We will use this later in the course to write chemical equations,"},{"Start":"04:07.700 ","End":"04:12.100","Text":"to describe chemical reactions."},{"Start":"04:16.630 ","End":"04:22.010","Text":"The second part of Dalton\u0027s atomic theory is that all atoms of"},{"Start":"04:22.010 ","End":"04:27.665","Text":"an element are identical but are different from the atoms of all other elements."},{"Start":"04:27.665 ","End":"04:32.430","Text":"This is the basis of the periodic table,"},{"Start":"04:36.170 ","End":"04:42.150","Text":"which lists all the elements that exist."},{"Start":"04:42.150 ","End":"04:46.385","Text":"We\u0027ll meet this in great detail later in the course."},{"Start":"04:46.385 ","End":"04:51.560","Text":"The third part of Dalton\u0027s atomic theory is that in each of their compounds,"},{"Start":"04:51.560 ","End":"04:55.865","Text":"different elements combine in a simple numerical ratio."},{"Start":"04:55.865 ","End":"04:58.880","Text":"Let\u0027s illustrate this with an example."},{"Start":"04:58.880 ","End":"05:02.825","Text":"Once again, we\u0027ll take the example of ascorbic acid,"},{"Start":"05:02.825 ","End":"05:04.760","Text":"which is vitamin C,"},{"Start":"05:04.760 ","End":"05:11.330","Text":"and its formula we said before is C_6H_8O_6."},{"Start":"05:11.330 ","End":"05:13.535","Text":"Now if we look at this formula,"},{"Start":"05:13.535 ","End":"05:17.270","Text":"we\u0027ll see that for every 6 carbon atoms,"},{"Start":"05:17.270 ","End":"05:19.790","Text":"there are 8 hydrogen atoms,"},{"Start":"05:19.790 ","End":"05:25.740","Text":"which means that the ratio of carbon to hydrogen is 6:8."},{"Start":"05:26.810 ","End":"05:30.385","Text":"We can simplify that to 3:4,"},{"Start":"05:30.385 ","End":"05:33.965","Text":"which is indeed a simple numerical ratio."},{"Start":"05:33.965 ","End":"05:39.110","Text":"We can go further than that and compare carbon to oxygen."},{"Start":"05:39.110 ","End":"05:43.940","Text":"We see there are 6 carbons for every 6 oxygens,"},{"Start":"05:43.940 ","End":"05:47.750","Text":"so the ratio is 6:6,"},{"Start":"05:47.750 ","End":"05:51.470","Text":"which is precisely the same as 1:1."},{"Start":"05:51.470 ","End":"05:56.430","Text":"In other words, another simple numerical ratio."},{"Start":"05:56.600 ","End":"05:58.620","Text":"Dalton was a very,"},{"Start":"05:58.620 ","End":"05:59.865","Text":"very good scientists,"},{"Start":"05:59.865 ","End":"06:03.315","Text":"so he used his theory to propose"},{"Start":"06:03.315 ","End":"06:08.205","Text":"a new law and he called it the law of multiple proportions."},{"Start":"06:08.205 ","End":"06:13.874","Text":"This new law says that if 2 elements for more than 1 compound,"},{"Start":"06:13.874 ","End":"06:16.639","Text":"for a fixed mass of 1 element,"},{"Start":"06:16.639 ","End":"06:19.280","Text":"the masses of the other elements in the compounds"},{"Start":"06:19.280 ","End":"06:21.964","Text":"will be in the ratio of small whole numbers."},{"Start":"06:21.964 ","End":"06:26.395","Text":"This says that if 2 elements from more than 1 compound,"},{"Start":"06:26.395 ","End":"06:28.885","Text":"for a fixed mass of 1 element,"},{"Start":"06:28.885 ","End":"06:31.850","Text":"the masses of the other elements in the compounds"},{"Start":"06:31.850 ","End":"06:35.675","Text":"will be in the ratio of small whole numbers."},{"Start":"06:35.675 ","End":"06:41.329","Text":"The example I\u0027ve chosen is carbon monoxide and carbon dioxide."},{"Start":"06:41.329 ","End":"06:46.055","Text":"Now both of these are compounds of carbon and oxygen,"},{"Start":"06:46.055 ","End":"06:48.395","Text":"but in different ratios."},{"Start":"06:48.395 ","End":"06:51.230","Text":"Let\u0027s first look at carbon monoxide."},{"Start":"06:51.230 ","End":"06:55.535","Text":"If we take 1 gram of carbon from carbon monoxide,"},{"Start":"06:55.535 ","End":"06:59.015","Text":"we\u0027ll discover that for every 1 gram of carbon,"},{"Start":"06:59.015 ","End":"07:01.945","Text":"there is 1.33 grams of oxygen."},{"Start":"07:01.945 ","End":"07:03.255","Text":"On the other hand,"},{"Start":"07:03.255 ","End":"07:06.315","Text":"let\u0027s look at the carbon dioxide."},{"Start":"07:06.315 ","End":"07:13.280","Text":"There, we see that for 1 gram of carbon there are 2.66 grams of oxygen."},{"Start":"07:13.280 ","End":"07:18.870","Text":"The ratio is 1.33:2.66,"},{"Start":"07:18.870 ","End":"07:21.190","Text":"which is precisely 1:2."},{"Start":"07:21.190 ","End":"07:26.750","Text":"Now there\u0027s a great difference between carbon monoxide and carbon dioxide."},{"Start":"07:26.750 ","End":"07:29.200","Text":"They\u0027re both colorless gases,"},{"Start":"07:29.200 ","End":"07:32.212","Text":"but the first one is highly poisonous."},{"Start":"07:32.212 ","End":"07:35.870","Text":"The second one is just the harmless substance,"},{"Start":"07:35.870 ","End":"07:39.860","Text":"the harmless gas you put into your fizzy drinks."},{"Start":"07:39.860 ","End":"07:47.495","Text":"Now we saw that the ratio was 1-2 and later on we\u0027ll see that for every carbon,"},{"Start":"07:47.495 ","End":"07:49.520","Text":"there is precisely 1 oxygen,"},{"Start":"07:49.520 ","End":"07:59.325","Text":"so the formula for carbon monoxide is written as CO. Mono in monoxide means 1."},{"Start":"07:59.325 ","End":"08:02.175","Text":"In carbon dioxide,"},{"Start":"08:02.175 ","End":"08:05.655","Text":"for every carbon, there are 2 oxygens."},{"Start":"08:05.655 ","End":"08:09.375","Text":"Di just means 2."},{"Start":"08:09.375 ","End":"08:17.390","Text":"Nowadays, we would write carbon monoxide as CO and carbon dioxide as CO_2."},{"Start":"08:17.390 ","End":"08:20.660","Text":"In this video, we\u0027ve discussed the laws that led to"},{"Start":"08:20.660 ","End":"08:25.310","Text":"Dalton\u0027s atomic theory and the law he derived from his theory."},{"Start":"08:25.310 ","End":"08:26.720","Text":"In the next video,"},{"Start":"08:26.720 ","End":"08:30.120","Text":"we\u0027ll discuss the structure of atoms."}],"ID":17641},{"Watched":false,"Name":"Exercise 1","Duration":"2m 44s","ChapterTopicVideoID":16883,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16883.jpeg","UploadDate":"2019-02-19T05:41:18.2830000","DurationForVideoObject":"PT2M44S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.195","Text":"Hi, we are going to answer the following question."},{"Start":"00:03.195 ","End":"00:07.020","Text":"When a solid mixture consisting of 10.75 grams calcium"},{"Start":"00:07.020 ","End":"00:11.850","Text":"hydroxide and 11.20 grams ammonium chloride is strongly heated,"},{"Start":"00:11.850 ","End":"00:17.070","Text":"gaseous products are evolved and 15.15 grams of a solid residue remains."},{"Start":"00:17.070 ","End":"00:20.130","Text":"The gases are passed into 60.35 grams of"},{"Start":"00:20.130 ","End":"00:24.540","Text":"water and the mass of the resulting solution is 67.15 grams."},{"Start":"00:24.540 ","End":"00:28.605","Text":"Show that these data conform to the law of conservation of mass."},{"Start":"00:28.605 ","End":"00:31.380","Text":"The law of conservation of mass states that"},{"Start":"00:31.380 ","End":"00:33.720","Text":"the total mass of the substances present after"},{"Start":"00:33.720 ","End":"00:39.250","Text":"a chemical reaction equals the total mass of the substances before the reaction."},{"Start":"00:39.250 ","End":"00:42.410","Text":"Here, the masses of the substances before"},{"Start":"00:42.410 ","End":"00:47.299","Text":"the reaction equal the mass of the calcium hydroxide,"},{"Start":"00:47.299 ","End":"00:49.385","Text":"which is 10.75 grams,"},{"Start":"00:49.385 ","End":"00:52.580","Text":"plus the mass of ammonium chloride,"},{"Start":"00:52.580 ","End":"00:54.990","Text":"which is 11.20 grams."},{"Start":"00:56.660 ","End":"01:00.465","Text":"This comes to 21.95 grams."},{"Start":"01:00.465 ","End":"01:04.805","Text":"Now, if we look at the mass of the substances after the reaction,"},{"Start":"01:04.805 ","End":"01:11.090","Text":"we have gaseous products and solid residue remains."},{"Start":"01:11.090 ","End":"01:21.515","Text":"The solid residue which remains is 15.15 grams of solid plus the mass of the gases."},{"Start":"01:21.515 ","End":"01:23.930","Text":"That\u0027s the mass of the gases."},{"Start":"01:23.930 ","End":"01:25.740","Text":"The mass of the gases,"},{"Start":"01:25.740 ","End":"01:27.290","Text":"in order to find what it is,"},{"Start":"01:27.290 ","End":"01:29.330","Text":"we need to look at this part of the question."},{"Start":"01:29.330 ","End":"01:33.065","Text":"The gases are passed into 60.35 grams of water"},{"Start":"01:33.065 ","End":"01:37.565","Text":"and the mass of the resulting solution is 67.15 grams."},{"Start":"01:37.565 ","End":"01:44.915","Text":"The increase in the mass from 60.35 to 67.15 was because of the gases."},{"Start":"01:44.915 ","End":"01:50.495","Text":"The mass of the gaseous products equals 67.15 grams,"},{"Start":"01:50.495 ","End":"01:56.210","Text":"which is the solution of the gases plus water minus the mass of the water,"},{"Start":"01:56.210 ","End":"01:59.010","Text":"which is 60.35 grams."},{"Start":"02:00.260 ","End":"02:03.400","Text":"This equals 6.8 grams."},{"Start":"02:03.400 ","End":"02:08.240","Text":"The mass of the substances after the reaction equals 15.15 grams of the"},{"Start":"02:08.240 ","End":"02:14.990","Text":"solid plus the mass of the gaseous products,"},{"Start":"02:14.990 ","End":"02:19.700","Text":"which is 6.8 grams of the gaseous products."},{"Start":"02:19.700 ","End":"02:25.035","Text":"This equals 21.95 grams."},{"Start":"02:25.035 ","End":"02:31.460","Text":"Here, we see that the mass of the substances before the reaction equals 21.95 grams,"},{"Start":"02:31.460 ","End":"02:36.995","Text":"and the mass of the substances after the reaction also equals 21.95 grams."},{"Start":"02:36.995 ","End":"02:41.240","Text":"Therefore, this data conform to the law of conservation of mass."},{"Start":"02:41.240 ","End":"02:42.935","Text":"That is our final answer."},{"Start":"02:42.935 ","End":"02:45.360","Text":"Thank you very much for listening."}],"ID":17642},{"Watched":false,"Name":"Exercise 2","Duration":"5m 16s","ChapterTopicVideoID":16884,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16884.jpeg","UploadDate":"2019-02-19T05:41:59.7630000","DurationForVideoObject":"PT5M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.285","Text":"Hi, we are going to answer the following question."},{"Start":"00:03.285 ","End":"00:09.480","Text":"In one experiment, 2.5 grams of sodium was allowed to react with 10 grams chlorine."},{"Start":"00:09.480 ","End":"00:14.685","Text":"All the sodium was used up and 5.65 grams sodium chloride was produced."},{"Start":"00:14.685 ","End":"00:16.580","Text":"In a second experiment,"},{"Start":"00:16.580 ","End":"00:20.740","Text":"2.27 grams chlorine was allowed to react with 10 grams of sodium."},{"Start":"00:20.740 ","End":"00:26.010","Text":"All the chlorine was used up and 3.56 grams sodium chloride was produced."},{"Start":"00:26.010 ","End":"00:30.330","Text":"Show that these results are consistent with the law of constant composition."},{"Start":"00:30.330 ","End":"00:33.300","Text":"The law of constant composition states"},{"Start":"00:33.300 ","End":"00:36.029","Text":"that all samples of a compound have the same composition,"},{"Start":"00:36.029 ","End":"00:39.630","Text":"the same proportions by mass of the constituent elements."},{"Start":"00:39.630 ","End":"00:41.640","Text":"Meaning, that if in Experiment 1,"},{"Start":"00:41.640 ","End":"00:45.340","Text":"we have sodium and chlorine, and then Experiment 2,"},{"Start":"00:45.340 ","End":"00:49.850","Text":"we also have sodium and chlorine and it comes to sodium chloride,"},{"Start":"00:49.850 ","End":"00:54.050","Text":"if we compare the percent by mass of the sodium"},{"Start":"00:54.050 ","End":"00:58.505","Text":"or of the chlorine in the sodium chloride, they should be equal."},{"Start":"00:58.505 ","End":"01:01.895","Text":"Let\u0027s begin. We have Experiment 1."},{"Start":"01:01.895 ","End":"01:03.785","Text":"We\u0027ll call it 1."},{"Start":"01:03.785 ","End":"01:10.385","Text":"In this experiment, the mass of the sodium equals to 0.05."},{"Start":"01:10.385 ","End":"01:14.030","Text":"This is reacted with an excess of chlorine."},{"Start":"01:14.030 ","End":"01:18.365","Text":"What we receive is 5.65 grams of sodium chloride."},{"Start":"01:18.365 ","End":"01:24.540","Text":"The mass of sodium chloride equals 5.65."},{"Start":"01:24.540 ","End":"01:30.509","Text":"Then second experiment, we have 2.27 grams of chlorine."},{"Start":"01:30.640 ","End":"01:37.650","Text":"The mass of chlorine equals 2.27 grams."},{"Start":"01:38.120 ","End":"01:43.240","Text":"The chlorine is reacted with an excess of sodium."},{"Start":"01:43.240 ","End":"01:47.030","Text":"We received 3.56 grams of sodium chloride."},{"Start":"01:47.030 ","End":"01:53.675","Text":"The mass of sodium chloride equals 3.56 grams."},{"Start":"01:53.675 ","End":"01:57.080","Text":"Here, we have the mass of the sodium and the mass"},{"Start":"01:57.080 ","End":"01:59.975","Text":"of the sodium chloride in Experiment 1."},{"Start":"01:59.975 ","End":"02:02.905","Text":"In order to find the mass of the chlorine,"},{"Start":"02:02.905 ","End":"02:05.420","Text":"we\u0027ll take the mass of the sodium chloride,"},{"Start":"02:05.420 ","End":"02:07.880","Text":"which is 5.65 grams,"},{"Start":"02:07.880 ","End":"02:11.360","Text":"minus the mass of the sodium,"},{"Start":"02:11.360 ","End":"02:13.680","Text":"which is 2.05 grams."},{"Start":"02:13.680 ","End":"02:18.135","Text":"This comes to 3.6 grams of chlorine."},{"Start":"02:18.135 ","End":"02:20.140","Text":"In Experiment 2,"},{"Start":"02:20.140 ","End":"02:22.970","Text":"the mass of the chlorine is 2.27 grams,"},{"Start":"02:22.970 ","End":"02:26.630","Text":"the mass of the sodium chloride is 3.56 grams."},{"Start":"02:26.630 ","End":"02:30.080","Text":"In order to find the mass of sodium,"},{"Start":"02:30.080 ","End":"02:32.240","Text":"we\u0027ll take the mass of the sodium chloride,"},{"Start":"02:32.240 ","End":"02:34.835","Text":"which is 3.56 grams,"},{"Start":"02:34.835 ","End":"02:37.250","Text":"minus the mass of chlorine,"},{"Start":"02:37.250 ","End":"02:39.955","Text":"which is 2.27 grams."},{"Start":"02:39.955 ","End":"02:44.820","Text":"The mass of the sodium comes to 1.29 grams."},{"Start":"02:44.820 ","End":"02:48.120","Text":"Now, we have Experiment 1 and Experiment 2, and in both of them,"},{"Start":"02:48.120 ","End":"02:49.430","Text":"we have the mass of the sodium,"},{"Start":"02:49.430 ","End":"02:50.840","Text":"mass of the chlorine,"},{"Start":"02:50.840 ","End":"02:52.565","Text":"and the mass of sodium chloride."},{"Start":"02:52.565 ","End":"02:56.825","Text":"In order to check if the results are consistent with the law of constant composition,"},{"Start":"02:56.825 ","End":"02:58.700","Text":"we will choose one of the elements,"},{"Start":"02:58.700 ","End":"03:00.685","Text":"sodium or chlorine."},{"Start":"03:00.685 ","End":"03:02.595","Text":"We\u0027ll choose sodium."},{"Start":"03:02.595 ","End":"03:04.781","Text":"We\u0027re going to see the percent by mass of sodium in"},{"Start":"03:04.781 ","End":"03:08.690","Text":"Experiment 1 and percent by mass of sodium in Experiment 2."},{"Start":"03:08.690 ","End":"03:14.435","Text":"If they are equal, then the results are consistent with the law of constant composition."},{"Start":"03:14.435 ","End":"03:15.785","Text":"In Experiment 1,"},{"Start":"03:15.785 ","End":"03:18.350","Text":"in order to find the percent by mass of the sodium,"},{"Start":"03:18.350 ","End":"03:19.670","Text":"we\u0027ll take the mass of the sodium,"},{"Start":"03:19.670 ","End":"03:22.315","Text":"which is 2.05 grams,"},{"Start":"03:22.315 ","End":"03:25.370","Text":"divided by the mass of sodium chloride,"},{"Start":"03:25.370 ","End":"03:32.820","Text":"which is 5.65 grams, times 100."},{"Start":"03:33.020 ","End":"03:38.325","Text":"This comes to 36.28 percent."},{"Start":"03:38.325 ","End":"03:40.570","Text":"Now, if we look at Experiment 2,"},{"Start":"03:40.570 ","End":"03:42.500","Text":"we have the mass of the sodium,"},{"Start":"03:42.500 ","End":"03:45.180","Text":"which is 1.29 grams,"},{"Start":"03:45.910 ","End":"03:49.490","Text":"divided by the mass of the sodium chloride,"},{"Start":"03:49.490 ","End":"03:57.260","Text":"which is 3.56 grams, times 100."},{"Start":"03:57.260 ","End":"04:02.500","Text":"This comes to 36.23 percent."},{"Start":"04:02.500 ","End":"04:05.160","Text":"As you can see in Experiment 1,"},{"Start":"04:05.160 ","End":"04:11.120","Text":"the percent by mass of sodium equals 36.28 percent."},{"Start":"04:11.120 ","End":"04:18.245","Text":"In Experiment 2, the percent by mass of sodium equals 36.23 percent. These are equal."},{"Start":"04:18.245 ","End":"04:22.895","Text":"Therefore, the results are consistent with the law of constant composition."},{"Start":"04:22.895 ","End":"04:25.385","Text":"Now, it\u0027s important for me to say two things."},{"Start":"04:25.385 ","End":"04:29.990","Text":"First of all, we chose to check the percent by mass of sodium."},{"Start":"04:29.990 ","End":"04:32.570","Text":"However, we could\u0027ve checked the percent by mass of"},{"Start":"04:32.570 ","End":"04:35.000","Text":"chlorine and it would\u0027ve been the same,"},{"Start":"04:35.000 ","End":"04:38.675","Text":"it would have been equal in Experiment 1 and in Experiment 2."},{"Start":"04:38.675 ","End":"04:40.670","Text":"Another important thing is,"},{"Start":"04:40.670 ","End":"04:43.625","Text":"since we have 2 elements in sodium chloride,"},{"Start":"04:43.625 ","End":"04:46.400","Text":"it\u0027s enough to check the percent by mass of just one element,"},{"Start":"04:46.400 ","End":"04:47.945","Text":"either the sodium or either chlorine,"},{"Start":"04:47.945 ","End":"04:51.890","Text":"because the total percent is 100 percent."},{"Start":"04:51.890 ","End":"04:55.790","Text":"For example, we got for sodium, around 36 percent."},{"Start":"04:55.790 ","End":"04:58.620","Text":"For chlorine, we will get around 64 percent."},{"Start":"04:58.620 ","End":"05:02.540","Text":"However, if we\u0027re checking molecules which have more than two elements, for example,"},{"Start":"05:02.540 ","End":"05:07.275","Text":"3 elements, we have to check the percent by mass of 2 elements and so on."},{"Start":"05:07.275 ","End":"05:08.630","Text":"This is our final answer."},{"Start":"05:08.630 ","End":"05:11.614","Text":"The results are consistent with the law of constant composition."},{"Start":"05:11.614 ","End":"05:13.445","Text":"Thank you very much for listening."},{"Start":"05:13.445 ","End":"05:16.530","Text":"We hope to hear from you again."}],"ID":17643},{"Watched":false,"Name":"Exercise 3","Duration":"2m 25s","ChapterTopicVideoID":16885,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16885.jpeg","UploadDate":"2019-02-19T05:42:29.6870000","DurationForVideoObject":"PT2M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.135","Text":"Hi. We are going to answer the following question."},{"Start":"00:03.135 ","End":"00:07.605","Text":"In one experiment, the burning of 0.336 grams of sulfur produced"},{"Start":"00:07.605 ","End":"00:12.735","Text":"0.645 grams of sulfur dioxide as a sole product of the reaction."},{"Start":"00:12.735 ","End":"00:14.355","Text":"In a second experiment,"},{"Start":"00:14.355 ","End":"00:17.535","Text":"0.856 grams sulfur dioxide was obtained."},{"Start":"00:17.535 ","End":"00:20.805","Text":"What mass of sulfur was burned in the second experiment?"},{"Start":"00:20.805 ","End":"00:26.050","Text":"In the first experiment, we know that the mass of sulfur is 0.336 grams."},{"Start":"00:29.000 ","End":"00:35.490","Text":"We also know that the mass of sulfur dioxide produced is 0.645 grams."},{"Start":"00:41.320 ","End":"00:49.470","Text":"Therefore, we know that the proportion of sulfur to sulfur dioxide equals 0.336 grams of"},{"Start":"00:49.470 ","End":"00:57.985","Text":"sulfur divided by 0.645 grams of sulfur dioxide."},{"Start":"00:57.985 ","End":"01:00.200","Text":"That\u0027s in the first experiment."},{"Start":"01:00.200 ","End":"01:02.435","Text":"In the second experiment,"},{"Start":"01:02.435 ","End":"01:07.770","Text":"the mass of sulfur dioxide is given and is 0.856 grams."},{"Start":"01:12.140 ","End":"01:18.560","Text":"We\u0027re asked to find the mass of sulfur burned in the second experiment."},{"Start":"01:18.560 ","End":"01:23.075","Text":"Since also in Experiment 1 and in Experiment 2,"},{"Start":"01:23.075 ","End":"01:25.655","Text":"we\u0027re talking about sulfur dioxide,"},{"Start":"01:25.655 ","End":"01:29.960","Text":"the proportion of sulfur to sulfur dioxide has to be equal in both samples."},{"Start":"01:29.960 ","End":"01:34.610","Text":"So in order to find the mass of sulfur in the second experiment,"},{"Start":"01:34.610 ","End":"01:40.250","Text":"we\u0027ll take the mass of sulfur dioxide and multiply it"},{"Start":"01:40.250 ","End":"01:45.680","Text":"by the proportion of sulfur to sulfur dioxide from the first experiment,"},{"Start":"01:45.680 ","End":"01:50.830","Text":"so times 0.336 grams of sulfur."},{"Start":"01:50.830 ","End":"01:53.960","Text":"Sorry, this was grams of sulfur dioxide."},{"Start":"01:53.960 ","End":"02:02.130","Text":"Grams of sulfur divided by 0.645 grams of sulfur dioxide."},{"Start":"02:03.130 ","End":"02:07.070","Text":"The grams of sulfur dioxide cancels out,"},{"Start":"02:07.070 ","End":"02:15.455","Text":"and our answer comes out to 0.446 grams of sulfur."},{"Start":"02:15.455 ","End":"02:21.530","Text":"The mass of sulfur from the second experiment equals 0.446 grams of sulfur."},{"Start":"02:21.530 ","End":"02:23.120","Text":"That is our final answer."},{"Start":"02:23.120 ","End":"02:25.620","Text":"Thank you very much for listening"}],"ID":17644},{"Watched":false,"Name":"Exercise 4","Duration":"3m 31s","ChapterTopicVideoID":24387,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24387.jpeg","UploadDate":"2021-03-13T03:24:17.3000000","DurationForVideoObject":"PT3M31S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.715","Text":"Hi. We are going to answer the following question."},{"Start":"00:02.715 ","End":"00:07.020","Text":"For the atom O-16 with mass 15.9949u,"},{"Start":"00:07.020 ","End":"00:08.715","Text":"determine, a,"},{"Start":"00:08.715 ","End":"00:10.230","Text":"the number of protons,"},{"Start":"00:10.230 ","End":"00:12.495","Text":"neutrons, and electrons in the atom,"},{"Start":"00:12.495 ","End":"00:17.490","Text":"and b, the ratio of the mass of this atom to that of the atom C-12."},{"Start":"00:17.490 ","End":"00:19.785","Text":"We will begin with a."},{"Start":"00:19.785 ","End":"00:23.580","Text":"To find the number of protons in the oxygen atom,"},{"Start":"00:23.580 ","End":"00:25.830","Text":"we will look in the periodic table of elements,"},{"Start":"00:25.830 ","End":"00:30.160","Text":"and we will find that the atomic number of oxygen,"},{"Start":"00:30.830 ","End":"00:34.980","Text":"which equals the number of protons, equals 8."},{"Start":"00:34.980 ","End":"00:36.780","Text":"So there are 8 protons in oxygen."},{"Start":"00:36.780 ","End":"00:39.185","Text":"Next, we will look for the number of neutrons."},{"Start":"00:39.185 ","End":"00:41.433","Text":"Here we see it\u0027s written O-16,"},{"Start":"00:41.433 ","End":"00:46.470","Text":"16 is the mass number."},{"Start":"00:46.570 ","End":"00:50.615","Text":"The mass number equals the number of protons plus neutrons."},{"Start":"00:50.615 ","End":"00:57.560","Text":"In our case, the mass number which appears here in the upper left corner is 16."},{"Start":"00:57.560 ","End":"00:58.880","Text":"The mass number of 16,"},{"Start":"00:58.880 ","End":"00:59.930","Text":"equals number of photons,"},{"Start":"00:59.930 ","End":"01:04.685","Text":"which is 8 plus the number of neutrons so the number of neutrons equals 8."},{"Start":"01:04.685 ","End":"01:06.485","Text":"Now to find the number of electrons,"},{"Start":"01:06.485 ","End":"01:10.055","Text":"since O is a neutral atom and it\u0027s not an ion,"},{"Start":"01:10.055 ","End":"01:14.435","Text":"the number of electrons equals the number of protons, which equals 8."},{"Start":"01:14.435 ","End":"01:18.645","Text":"Therefore, we got the number of protons equals 8,"},{"Start":"01:18.645 ","End":"01:20.835","Text":"the number of neutrons equals 8,"},{"Start":"01:20.835 ","End":"01:25.245","Text":"and the number of electrons also equals 8."},{"Start":"01:25.245 ","End":"01:26.925","Text":"We will now go on to b."},{"Start":"01:26.925 ","End":"01:30.380","Text":"In b, we\u0027re asked to find the ratio of the mass of this atom,"},{"Start":"01:30.380 ","End":"01:33.110","Text":"which is given here, you see the mass,"},{"Start":"01:33.110 ","End":"01:36.295","Text":"to that of the atom C-12."},{"Start":"01:36.295 ","End":"01:40.310","Text":"We want to find the ratio of the mass of O-16,"},{"Start":"01:40.310 ","End":"01:42.890","Text":"which is given, the mess of C-12."},{"Start":"01:42.890 ","End":"01:48.300","Text":"Again, the mass of O-16 is given and equals 15.9949u."},{"Start":"01:52.330 ","End":"01:56.825","Text":"Now we want to divide it by the mass of 12C."},{"Start":"01:56.825 ","End":"02:01.910","Text":"Carbon-12 is arbitrarily assigned a mass of exactly 12 atomic units,"},{"Start":"02:01.910 ","End":"02:08.645","Text":"which is 12u so divided by 12u, equals 1.33."},{"Start":"02:08.645 ","End":"02:16.250","Text":"The ratio of the mass of the O-16 atom to that of the C-12 atom equals 1.33."},{"Start":"02:16.250 ","End":"02:17.540","Text":"That is our final answer."},{"Start":"02:17.540 ","End":"02:20.010","Text":"Thank you very much for listening."}],"ID":25220},{"Watched":false,"Name":"Atomic Structure","Duration":"8m 54s","ChapterTopicVideoID":16896,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16896.jpeg","UploadDate":"2019-02-19T05:47:09.0230000","DurationForVideoObject":"PT8M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:04.800","Text":"In this video, we will talk about the structure of the atom."},{"Start":"00:04.800 ","End":"00:11.190","Text":"Rutherford proposed a model of the atom called the nuclear atom in 1909."},{"Start":"00:11.190 ","End":"00:14.280","Text":"In his model which was based on experiments,"},{"Start":"00:14.280 ","End":"00:17.895","Text":"there are 3 particles: protons,"},{"Start":"00:17.895 ","End":"00:19.980","Text":"neutrons, and electrons."},{"Start":"00:19.980 ","End":"00:22.980","Text":"We\u0027ll indicate these by p, n,"},{"Start":"00:22.980 ","End":"00:27.750","Text":"and e. This theory proved to be correct and Rutherford"},{"Start":"00:27.750 ","End":"00:34.300","Text":"himself managed to identify protons in 1919."},{"Start":"00:34.880 ","End":"00:40.890","Text":"Neutrons were discovered many years later by Chadwick,"},{"Start":"00:40.890 ","End":"00:46.200","Text":"and it wasn\u0027t until 1932 that he managed to discover them."},{"Start":"00:46.200 ","End":"00:51.710","Text":"We have 3 particles inside the atom: proton,"},{"Start":"00:51.710 ","End":"00:54.655","Text":"neutron, and electron."},{"Start":"00:54.655 ","End":"01:00.695","Text":"Let\u0027s look at a simple diagram of the atom."},{"Start":"01:00.695 ","End":"01:05.100","Text":"Here\u0027s a schematic picture of an atom."},{"Start":"01:05.260 ","End":"01:08.330","Text":"In the center,"},{"Start":"01:08.330 ","End":"01:11.360","Text":"we have 2 particles."},{"Start":"01:11.360 ","End":"01:13.085","Text":"We have protons."},{"Start":"01:13.085 ","End":"01:15.320","Text":"These are the ones with the p on them,"},{"Start":"01:15.320 ","End":"01:17.540","Text":"with a plus on them,"},{"Start":"01:17.540 ","End":"01:20.330","Text":"and we have neutrons which don\u0027t have anything written on them."},{"Start":"01:20.330 ","End":"01:24.720","Text":"These together constitute the nucleus."},{"Start":"01:25.630 ","End":"01:34.140","Text":"They occupy very small area with a very high mass."},{"Start":"01:35.660 ","End":"01:38.085","Text":"Round about them,"},{"Start":"01:38.085 ","End":"01:39.875","Text":"there are electrons,"},{"Start":"01:39.875 ","End":"01:42.650","Text":"which follow some path."},{"Start":"01:42.650 ","End":"01:45.140","Text":"We don\u0027t know exactly which path,"},{"Start":"01:45.140 ","End":"01:47.105","Text":"but we\u0027ll discuss it later."},{"Start":"01:47.105 ","End":"01:51.440","Text":"The electrons are indicated here by the little minus signs."},{"Start":"01:51.440 ","End":"01:54.990","Text":"Now it\u0027s known that the distance between the nucleus,"},{"Start":"01:54.990 ","End":"01:56.525","Text":"which is the center,"},{"Start":"01:56.525 ","End":"02:00.050","Text":"and the electrons is very large."},{"Start":"02:00.050 ","End":"02:03.025","Text":"In between, them there is absolutely nothing."},{"Start":"02:03.025 ","End":"02:10.720","Text":"We can say that most of an atom is just free space."},{"Start":"02:13.330 ","End":"02:18.955","Text":"Now we\u0027re going to say a little about the charge of the particles."},{"Start":"02:18.955 ","End":"02:27.095","Text":"We saw before that I wrote the proton with a little positive sign on it."},{"Start":"02:27.095 ","End":"02:32.645","Text":"The proton charge is plus 1 atomic units of charge."},{"Start":"02:32.645 ","End":"02:36.740","Text":"We\u0027ll discuss afterwards what an atomic unit of charge is,"},{"Start":"02:36.740 ","End":"02:38.575","Text":"but it\u0027s plus 1."},{"Start":"02:38.575 ","End":"02:44.915","Text":"Often we just write that the proton charge is plus 1 or even just plus."},{"Start":"02:44.915 ","End":"02:50.542","Text":"The electron charge is minus 1 atomic units of charge,"},{"Start":"02:50.542 ","End":"02:55.075","Text":"exactly the opposite sign as the proton."},{"Start":"02:55.075 ","End":"02:59.915","Text":"We just often write it just as minus 1 or minus."},{"Start":"02:59.915 ","End":"03:02.525","Text":"The neutron is,"},{"Start":"03:02.525 ","End":"03:05.185","Text":"as the name suggests, completely neutral."},{"Start":"03:05.185 ","End":"03:07.644","Text":"It has a 0 charge."},{"Start":"03:07.644 ","End":"03:10.465","Text":"Now what is the atomic unit of charge?"},{"Start":"03:10.465 ","End":"03:16.540","Text":"It\u0027s 1.6022 times 10 to the minus 19 coulombs,"},{"Start":"03:16.540 ","End":"03:20.065","Text":"and that\u0027s a very small amount of charge."},{"Start":"03:20.065 ","End":"03:25.310","Text":"The coulomb is the SI unit for charge."},{"Start":"03:26.760 ","End":"03:30.340","Text":"You\u0027ll probably meet it in your physics classes."},{"Start":"03:30.340 ","End":"03:35.980","Text":"What we need to know at this stage is that the proton and the electron are precisely"},{"Start":"03:35.980 ","End":"03:38.740","Text":"the opposite charge and that their charge of"},{"Start":"03:38.740 ","End":"03:43.045","Text":"a single proton or a single electron is extremely small."},{"Start":"03:43.045 ","End":"03:45.835","Text":"Now, an atom is a neutral thing."},{"Start":"03:45.835 ","End":"03:47.350","Text":"In order to be neutral,"},{"Start":"03:47.350 ","End":"03:53.750","Text":"it has to have the same number of protons as number of electrons."},{"Start":"03:53.870 ","End":"03:59.200","Text":"Having said something about the charge,"},{"Start":"03:59.200 ","End":"04:02.260","Text":"we now say something about the mass of the particles."},{"Start":"04:02.260 ","End":"04:08.615","Text":"The mass of a proton is 1.0073 atomic mass units."},{"Start":"04:08.615 ","End":"04:12.780","Text":"We talked about atomic mass units when we talked about SI units."},{"Start":"04:12.780 ","End":"04:15.565","Text":"That number is very close to 1,"},{"Start":"04:15.565 ","End":"04:22.640","Text":"so we can write that the mass of the proton is approximately 1 atomic mass unit."},{"Start":"04:22.640 ","End":"04:27.080","Text":"The mass of the neutron is a very similar number to the mass of the proton."},{"Start":"04:27.080 ","End":"04:31.250","Text":"It\u0027s 1.0087 atomic mass units."},{"Start":"04:31.250 ","End":"04:34.265","Text":"That\u0027s also approximately 1."},{"Start":"04:34.265 ","End":"04:40.130","Text":"We can say that the mass of the proton is approximately equal to the mass of"},{"Start":"04:40.130 ","End":"04:45.500","Text":"the neutron and they are approximately 1 atomic mass unit."},{"Start":"04:45.500 ","End":"04:48.275","Text":"Let\u0027s look at the mass of the electron."},{"Start":"04:48.275 ","End":"04:56.540","Text":"The mass of the electron is 0.00054858 atomic mass units."},{"Start":"04:56.540 ","End":"04:59.874","Text":"That\u0027s almost negligible, almost 0."},{"Start":"04:59.874 ","End":"05:03.110","Text":"We can see that the mass of"},{"Start":"05:03.110 ","End":"05:07.924","Text":"an atom is determined by the mass of the protons and the mass of the neutrons,"},{"Start":"05:07.924 ","End":"05:11.575","Text":"far more than it\u0027s determined by the mass of the electrons."},{"Start":"05:11.575 ","End":"05:14.595","Text":"Now, what is 1 atomic mass unit?"},{"Start":"05:14.595 ","End":"05:19.460","Text":"It\u0027s 1.66 times 10 to the minus 27 kilograms."},{"Start":"05:19.460 ","End":"05:23.260","Text":"In other words, it\u0027s an extremely small mass."},{"Start":"05:23.260 ","End":"05:30.200","Text":"We would need to take many atoms together in order to form 1 kilogram."},{"Start":"05:30.200 ","End":"05:32.490","Text":"Each atom on its own weighs"},{"Start":"05:32.490 ","End":"05:37.090","Text":"very little [inaudible] any sample of"},{"Start":"05:37.090 ","End":"05:42.235","Text":"matter that we investigate will consist of a great many atoms."},{"Start":"05:42.235 ","End":"05:46.900","Text":"Later, we\u0027ll explain how the atomic mass unit is defined."},{"Start":"05:46.900 ","End":"05:52.180","Text":"Now, the 2 [inaudible] concepts that we have to explain."},{"Start":"05:52.180 ","End":"05:55.330","Text":"The atomic number is the first one."},{"Start":"05:55.330 ","End":"06:01.510","Text":"The atomic number which we indicate by the letter Z is the number of protons."},{"Start":"06:01.510 ","End":"06:07.830","Text":"Z is equal to the number of protons."},{"Start":"06:07.830 ","End":"06:11.330","Text":"Now, Z is a very important number"},{"Start":"06:11.330 ","End":"06:15.830","Text":"because every element is characterized by different number of protons."},{"Start":"06:15.830 ","End":"06:18.890","Text":"That\u0027s what distinguishes one element from another."},{"Start":"06:18.890 ","End":"06:22.374","Text":"Let\u0027s take examples. In hydrogen,"},{"Start":"06:22.374 ","End":"06:25.095","Text":"Z=1, that means there\u0027s 1 proton."},{"Start":"06:25.095 ","End":"06:29.325","Text":"In carbon, Z=6, that means there are 6 protons."},{"Start":"06:29.325 ","End":"06:31.275","Text":"In oxygen, Z=8,"},{"Start":"06:31.275 ","End":"06:33.705","Text":"that means there are 8 protons."},{"Start":"06:33.705 ","End":"06:35.840","Text":"Now, in a neutral atom,"},{"Start":"06:35.840 ","End":"06:40.175","Text":"the number of protons must be equal to the number of electrons."},{"Start":"06:40.175 ","End":"06:41.955","Text":"In a hydrogen atom,"},{"Start":"06:41.955 ","End":"06:43.695","Text":"there will be only 1 electron,"},{"Start":"06:43.695 ","End":"06:45.780","Text":"in a carbon atom,"},{"Start":"06:45.780 ","End":"06:47.355","Text":"only 6 electrons,"},{"Start":"06:47.355 ","End":"06:48.930","Text":"and in an oxygen, atom,"},{"Start":"06:48.930 ","End":"06:51.510","Text":"there will be 8 electrons."},{"Start":"06:51.510 ","End":"06:53.100","Text":"If the atom is neutral,"},{"Start":"06:53.100 ","End":"06:56.015","Text":"the number of electrons is also given by Z."},{"Start":"06:56.015 ","End":"06:58.535","Text":"In addition to atomic number,"},{"Start":"06:58.535 ","End":"07:00.125","Text":"we also have atomic mass."},{"Start":"07:00.125 ","End":"07:01.910","Text":"What\u0027s the atomic mass?"},{"Start":"07:01.910 ","End":"07:05.420","Text":"The atomic mass, which we indicate by the letter A,"},{"Start":"07:05.420 ","End":"07:08.600","Text":"is the sum of the number of protons and neutrons."},{"Start":"07:08.600 ","End":"07:15.985","Text":"A is equal to the number of protons plus the number of neutrons."},{"Start":"07:15.985 ","End":"07:19.715","Text":"It\u0027s called the atomic mass because as we saw before,"},{"Start":"07:19.715 ","End":"07:27.690","Text":"each proton and each neutron has a mass of approximately 1 atomic mass unit."},{"Start":"07:27.690 ","End":"07:29.790","Text":"If we add the 2 together,"},{"Start":"07:29.790 ","End":"07:36.740","Text":"A gives us an indication of the mass of the nucleus and the mass of the atom."},{"Start":"07:36.740 ","End":"07:46.230","Text":"Now, we said that A is the number of protons plus the number of neutrons,"},{"Start":"07:46.230 ","End":"07:50.905","Text":"and A minus Z is then the number of protons"},{"Start":"07:50.905 ","End":"07:56.370","Text":"plus the number of neutrons minus the number of protons."},{"Start":"07:56.370 ","End":"08:00.580","Text":"A minus Z gives us the number of neutrons."},{"Start":"08:00.580 ","End":"08:02.335","Text":"Let\u0027s have an example."},{"Start":"08:02.335 ","End":"08:04.435","Text":"Let\u0027s consider carbon."},{"Start":"08:04.435 ","End":"08:09.180","Text":"Carbon has Z=6 and A=12."},{"Start":"08:09.180 ","End":"08:12.270","Text":"As soon as you see that Z=6,"},{"Start":"08:12.270 ","End":"08:17.290","Text":"we know that the number of protons is equal to 6,"},{"Start":"08:17.290 ","End":"08:21.425","Text":"which means that the number of electrons must also be 6."},{"Start":"08:21.425 ","End":"08:26.220","Text":"The number of protons plus the number of neutrons is 12,"},{"Start":"08:26.220 ","End":"08:33.900","Text":"so the number of neutrons must be 12 minus 6 equal to 6."},{"Start":"08:33.900 ","End":"08:37.980","Text":"We know that carbon has 6 protons,"},{"Start":"08:37.980 ","End":"08:41.420","Text":"6 electrons, and 6 neutrons."},{"Start":"08:41.420 ","End":"08:46.055","Text":"In this video, we discussed the subatomic particles in an atom,"},{"Start":"08:46.055 ","End":"08:49.200","Text":"we also talked about their charge and mass,"},{"Start":"08:49.200 ","End":"08:54.300","Text":"and also about the atomic number and atomic mass."}],"ID":17645},{"Watched":false,"Name":"Elements Isotopes and Ions","Duration":"8m 19s","ChapterTopicVideoID":16897,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16897.jpeg","UploadDate":"2019-02-19T05:48:32.0300000","DurationForVideoObject":"PT8M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"In this video, we will discuss the symbols for atoms,"},{"Start":"00:04.290 ","End":"00:06.329","Text":"isotopes, and ions."},{"Start":"00:06.329 ","End":"00:07.650","Text":"In order to do this,"},{"Start":"00:07.650 ","End":"00:11.475","Text":"it\u0027s a good idea to remember some of the things we\u0027ve learned before."},{"Start":"00:11.475 ","End":"00:15.315","Text":"Each element has atoms with a particular value of Z."},{"Start":"00:15.315 ","End":"00:17.920","Text":"Z is the atomic number,"},{"Start":"00:18.590 ","End":"00:21.150","Text":"the number of protons."},{"Start":"00:21.150 ","End":"00:23.910","Text":"The second thing we need to remember is that the number of"},{"Start":"00:23.910 ","End":"00:27.750","Text":"electrons in a neutral atom is equal to the number protons,"},{"Start":"00:27.750 ","End":"00:30.165","Text":"so it\u0027s also the same as Z."},{"Start":"00:30.165 ","End":"00:36.185","Text":"The number of protons is equal to the number of electrons in a neutral atom."},{"Start":"00:36.185 ","End":"00:39.410","Text":"The third thing we need to remember is that the mass number"},{"Start":"00:39.410 ","End":"00:43.280","Text":"A is the total number of protons and neutrons."},{"Start":"00:43.280 ","End":"00:48.500","Text":"It\u0027s the number of protons plus the number of neutrons."},{"Start":"00:48.500 ","End":"00:51.260","Text":"Now, elements have historical names,"},{"Start":"00:51.260 ","End":"00:53.195","Text":"not just the value of Z."},{"Start":"00:53.195 ","End":"00:55.745","Text":"Let\u0027s take some examples."},{"Start":"00:55.745 ","End":"00:57.815","Text":"The first set of examples,"},{"Start":"00:57.815 ","End":"01:02.420","Text":"we\u0027ve taken are ones in which the symbol for them is rather obvious."},{"Start":"01:02.420 ","End":"01:06.995","Text":"For example, oxygen is symbolized by O,"},{"Start":"01:06.995 ","End":"01:08.795","Text":"carbon by C,"},{"Start":"01:08.795 ","End":"01:10.640","Text":"phosphorus with P,"},{"Start":"01:10.640 ","End":"01:14.705","Text":"sulfur with S, selenium with Se."},{"Start":"01:14.705 ","End":"01:19.275","Text":"We couldn\u0027t use S because S had already been used for sulfur,"},{"Start":"01:19.275 ","End":"01:21.980","Text":"so you just add the second letter of selenium."},{"Start":"01:21.980 ","End":"01:25.030","Text":"Likewise, neon is N,"},{"Start":"01:25.030 ","End":"01:29.150","Text":"N and e for the second letter of neon."},{"Start":"01:29.150 ","End":"01:33.874","Text":"The reason we can\u0027t use N alone is because N has been used for nitrogen."},{"Start":"01:33.874 ","End":"01:36.170","Text":"Now, the next group are less"},{"Start":"01:36.170 ","End":"01:39.725","Text":"obvious unless you happen to know Latin or Greek rather well."},{"Start":"01:39.725 ","End":"01:44.150","Text":"The first one is potassium and that\u0027s symbolized by the K,"},{"Start":"01:44.150 ","End":"01:47.565","Text":"sodium by N and a,"},{"Start":"01:47.565 ","End":"01:50.475","Text":"silver by A and g,"},{"Start":"01:50.475 ","End":"01:52.890","Text":"gold by A and u,"},{"Start":"01:52.890 ","End":"01:55.185","Text":"lead by P and b."},{"Start":"01:55.185 ","End":"02:00.860","Text":"Final one, lead might be quite familiar because it comes from the Latin plumbum,"},{"Start":"02:00.860 ","End":"02:03.260","Text":"which is also the root of plumber."},{"Start":"02:03.260 ","End":"02:05.660","Text":"That\u0027s because at one time,"},{"Start":"02:05.660 ","End":"02:09.290","Text":"pipes that carry water were made of lead."},{"Start":"02:09.290 ","End":"02:14.765","Text":"Each atom of a particular element can come in several varieties."},{"Start":"02:14.765 ","End":"02:17.210","Text":"They always have the same number of protons,"},{"Start":"02:17.210 ","End":"02:21.365","Text":"but they can have different numbers of neutrons."},{"Start":"02:21.365 ","End":"02:25.595","Text":"We write them with E at the center,"},{"Start":"02:25.595 ","End":"02:27.710","Text":"the volume of Z at the bottom,"},{"Start":"02:27.710 ","End":"02:30.520","Text":"an A on the top."},{"Start":"02:30.520 ","End":"02:32.905","Text":"Let\u0027s take some examples."},{"Start":"02:32.905 ","End":"02:35.305","Text":"The example I\u0027ve chosen is carbon."},{"Start":"02:35.305 ","End":"02:39.620","Text":"Now, for carbon, Z is always = 6."},{"Start":"02:39.620 ","End":"02:46.050","Text":"We can write 3 versions of carbon, all with Z=6."},{"Start":"02:46.050 ","End":"02:49.015","Text":"All of them have 6 protons,"},{"Start":"02:49.015 ","End":"02:52.450","Text":"but they have different numbers of neutrons."},{"Start":"02:52.450 ","End":"02:56.215","Text":"The first one has 6 neutrons, the second,"},{"Start":"02:56.215 ","End":"03:00.205","Text":"7 neutrons, and the third, 8 neutrons."},{"Start":"03:00.205 ","End":"03:04.645","Text":"If we add the number of protons and the number of neutrons to get the total,"},{"Start":"03:04.645 ","End":"03:06.230","Text":"we get 12,"},{"Start":"03:06.230 ","End":"03:09.285","Text":"here\u0027s 13, and here 14."},{"Start":"03:09.285 ","End":"03:12.670","Text":"These are the values of A,"},{"Start":"03:12.670 ","End":"03:14.860","Text":"the atomic mass number."},{"Start":"03:14.860 ","End":"03:19.720","Text":"Now we can write here 12, 13, 14."},{"Start":"03:19.720 ","End":"03:23.045","Text":"These are 3 different isotopes of carbon."},{"Start":"03:23.045 ","End":"03:25.325","Text":"We refer to as carbon-12,"},{"Start":"03:25.325 ","End":"03:27.965","Text":"carbon-13, and carbon-14."},{"Start":"03:27.965 ","End":"03:30.425","Text":"You\u0027ve probably all heard of carbon-14,"},{"Start":"03:30.425 ","End":"03:34.960","Text":"because it\u0027s used to date substances that contain carbon."},{"Start":"03:34.960 ","End":"03:38.945","Text":"Now, isotopes are not all equally abundant."},{"Start":"03:38.945 ","End":"03:42.110","Text":"Sometimes you have more of one isotope than another."},{"Start":"03:42.110 ","End":"03:46.085","Text":"The one we\u0027re going to choose to talk about is once again carbon."},{"Start":"03:46.085 ","End":"03:51.430","Text":"For carbon, carbon-12 is the most abundant, carbon-13,"},{"Start":"03:51.430 ","End":"03:53.055","Text":"far less abundant,"},{"Start":"03:53.055 ","End":"03:56.420","Text":"and even less abundant is carbon-14,"},{"Start":"03:56.420 ","End":"03:58.970","Text":"which is good because it\u0027s radioactive."},{"Start":"03:58.970 ","End":"04:03.755","Text":"So far, we\u0027ve talked about elements and atoms as being neutral."},{"Start":"04:03.755 ","End":"04:10.920","Text":"That means they have the same number of protons as electrons."},{"Start":"04:10.990 ","End":"04:20.465","Text":"Now, ions occur when there is an unequal amount of protons and electrons."},{"Start":"04:20.465 ","End":"04:25.268","Text":"We can either have more protons than electrons or more electrons than protons."},{"Start":"04:25.268 ","End":"04:31.085","Text":"Let\u0027s take the example first where we have more electrons than protons."},{"Start":"04:31.085 ","End":"04:36.470","Text":"Now, if we have more electrons than protons,"},{"Start":"04:38.600 ","End":"04:42.080","Text":"remember that electrons are negatively charged,"},{"Start":"04:42.080 ","End":"04:44.075","Text":"protons are positively charged."},{"Start":"04:44.075 ","End":"04:46.264","Text":"If there are more electrons than protons,"},{"Start":"04:46.264 ","End":"04:49.255","Text":"the overall charge will be negative,"},{"Start":"04:49.255 ","End":"04:52.570","Text":"and this we call an anion."},{"Start":"04:52.570 ","End":"04:57.810","Text":"Now, if we have more protons than electrons,"},{"Start":"04:58.810 ","End":"05:02.330","Text":"because electrons have been lost,"},{"Start":"05:02.330 ","End":"05:06.290","Text":"then this will be positively charged because the protons are positively"},{"Start":"05:06.290 ","End":"05:11.450","Text":"charged and there are more of them than there are of the negatively charged electrons."},{"Start":"05:11.450 ","End":"05:18.420","Text":"These are called cations and they are positively charged."},{"Start":"05:18.730 ","End":"05:21.155","Text":"We can make a rule."},{"Start":"05:21.155 ","End":"05:25.980","Text":"The rule is that the charge in an ion is equal to the number of protons,"},{"Start":"05:25.980 ","End":"05:27.875","Text":"which are positively charged,"},{"Start":"05:27.875 ","End":"05:29.660","Text":"minus the number of electrons,"},{"Start":"05:29.660 ","End":"05:31.775","Text":"which are negatively charged."},{"Start":"05:31.775 ","End":"05:33.995","Text":"Let\u0027s take an example."},{"Start":"05:33.995 ","End":"05:40.925","Text":"The first example we\u0027re going to take is an anion of oxygen."},{"Start":"05:40.925 ","End":"05:45.215","Text":"Ordinary oxygen, the most common isotope of oxygen,"},{"Start":"05:45.215 ","End":"05:48.675","Text":"is this one, oxygen-16."},{"Start":"05:48.675 ","End":"05:51.120","Text":"It has 8 protons,"},{"Start":"05:51.120 ","End":"05:55.755","Text":"8 neutrons, and 8 electrons."},{"Start":"05:55.755 ","End":"06:01.950","Text":"Now, if it acquires 2 electrons,"},{"Start":"06:01.950 ","End":"06:06.475","Text":"that\u0027s another 2, it will have 10 electrons."},{"Start":"06:06.475 ","End":"06:13.160","Text":"We can write the charge as the number of protons minus the number of electrons."},{"Start":"06:13.160 ","End":"06:15.410","Text":"That\u0027s the number of protons is 8,"},{"Start":"06:15.410 ","End":"06:17.525","Text":"the number of electrons is 10,"},{"Start":"06:17.525 ","End":"06:21.990","Text":"8 minus 10 is equal to minus 2,"},{"Start":"06:21.990 ","End":"06:24.530","Text":"and we\u0027ll write it 2 minus."},{"Start":"06:24.530 ","End":"06:27.575","Text":"This is the anion of oxygen."},{"Start":"06:27.575 ","End":"06:35.535","Text":"Now magnesium has a very common isotope called magnesium-24."},{"Start":"06:35.535 ","End":"06:38.445","Text":"Magnesium is Z=2,"},{"Start":"06:38.445 ","End":"06:41.965","Text":"and this isotope has an A=24."},{"Start":"06:41.965 ","End":"06:44.530","Text":"That means it has 12 protons,"},{"Start":"06:44.530 ","End":"06:49.555","Text":"12 neutrons, and it has 12 electrons."},{"Start":"06:49.555 ","End":"06:52.365","Text":"Now, if it loses 2 electrons,"},{"Start":"06:52.365 ","End":"06:55.680","Text":"minus 2 electrons, it will have,"},{"Start":"06:55.680 ","End":"06:57.895","Text":"again, 10 electrons."},{"Start":"06:57.895 ","End":"07:00.355","Text":"The total charge will be 12,"},{"Start":"07:00.355 ","End":"07:03.490","Text":"the number of protons, minus 10,"},{"Start":"07:03.490 ","End":"07:04.975","Text":"the number of electrons,"},{"Start":"07:04.975 ","End":"07:07.060","Text":"giving us plus 2,"},{"Start":"07:07.060 ","End":"07:10.280","Text":"which we write here, 2 plus."},{"Start":"07:10.280 ","End":"07:15.900","Text":"This is a common isotope of magnesium, magnesium^2 plus."},{"Start":"07:15.900 ","End":"07:18.715","Text":"So this is a cation,"},{"Start":"07:18.715 ","End":"07:22.795","Text":"whereas the one we discussed of oxygen was an anion."},{"Start":"07:22.795 ","End":"07:25.825","Text":"You can notice several interesting things."},{"Start":"07:25.825 ","End":"07:27.820","Text":"You notice the number of electrons,"},{"Start":"07:27.820 ","End":"07:31.480","Text":"10 electrons here is equal to 10 electrons here."},{"Start":"07:31.480 ","End":"07:37.720","Text":"These 2 ions are said to be isoelectronic,"},{"Start":"07:37.720 ","End":"07:40.899","Text":"the same number of electron."},{"Start":"07:40.899 ","End":"07:44.375","Text":"In addition, we can see one other thing."},{"Start":"07:44.375 ","End":"07:51.780","Text":"We can see that the magnesium^2 plus is the cation of a metal,"},{"Start":"07:51.780 ","End":"07:54.160","Text":"magnesium is a metal."},{"Start":"07:54.590 ","End":"08:03.165","Text":"Whereas oxygen, which has anion O^2 minus is a non-metal."},{"Start":"08:03.165 ","End":"08:05.390","Text":"That\u0027s generally the truth."},{"Start":"08:05.390 ","End":"08:11.285","Text":"Metals give cations and non-metals give anions."},{"Start":"08:11.285 ","End":"08:15.095","Text":"In this video, we\u0027ve learned to write the symbols for elements,"},{"Start":"08:15.095 ","End":"08:18.390","Text":"their isotopes, and ions."}],"ID":17646},{"Watched":false,"Name":"Exercise 5","Duration":"2m 23s","ChapterTopicVideoID":24388,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24388.jpeg","UploadDate":"2021-03-13T03:24:22.6230000","DurationForVideoObject":"PT2M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:03.045","Text":"Hey, we\u0027re going to answer the following question."},{"Start":"00:03.045 ","End":"00:05.685","Text":"There are 2 naturally occurring isotopes of bromine."},{"Start":"00:05.685 ","End":"00:08.940","Text":"They have the following abundances: bromine 79,"},{"Start":"00:08.940 ","End":"00:14.070","Text":"50.69 percent and bromine 81, 49.31 percent."},{"Start":"00:14.070 ","End":"00:19.080","Text":"The mass of bromine 79 is 78.918336u,"},{"Start":"00:19.080 ","End":"00:21.225","Text":"what is the mass of bromine 81?"},{"Start":"00:21.225 ","End":"00:22.860","Text":"In order to solve this question,"},{"Start":"00:22.860 ","End":"00:24.750","Text":"we will use the formula,"},{"Start":"00:24.750 ","End":"00:27.840","Text":"atomic mass equals the mass of the first isotope,"},{"Start":"00:27.840 ","End":"00:34.240","Text":"so the mass of isotope 1 times the fractional abundance of the first isotope."},{"Start":"00:34.730 ","End":"00:37.905","Text":"Fractional abundance of isotope 1,"},{"Start":"00:37.905 ","End":"00:41.019","Text":"plus the mass of the second isotope,"},{"Start":"00:41.660 ","End":"00:47.510","Text":"times the fractional abundance of the second isotope of isotope 2."},{"Start":"00:47.510 ","End":"00:48.650","Text":"And so, for example,"},{"Start":"00:48.650 ","End":"00:49.790","Text":"if we have 3 isotopes,"},{"Start":"00:49.790 ","End":"00:51.115","Text":"we\u0027ll have 3 contributions,"},{"Start":"00:51.115 ","End":"00:53.165","Text":"we\u0027ll have another contribution here."},{"Start":"00:53.165 ","End":"00:58.275","Text":"In our case, we are looking for the mass of bromine 81."},{"Start":"00:58.275 ","End":"01:00.380","Text":"We will start with the atomic mass."},{"Start":"01:00.380 ","End":"01:01.670","Text":"The atomic mass is found in"},{"Start":"01:01.670 ","End":"01:04.130","Text":"the periodic table of elements and if we will look at bromine,"},{"Start":"01:04.130 ","End":"01:10.855","Text":"we will see that the atomic mass equals 79.904u,"},{"Start":"01:10.855 ","End":"01:12.050","Text":"because it\u0027s atomic units,"},{"Start":"01:12.050 ","End":"01:14.380","Text":"equals the mass of isotope 1,"},{"Start":"01:14.380 ","End":"01:19.435","Text":"which is given in here, 78.918336u."},{"Start":"01:19.435 ","End":"01:26.990","Text":"78.918336u times the fractional abundance of the isotope."},{"Start":"01:26.990 ","End":"01:30.170","Text":"The abundance of the isotope is 50.69 percent."},{"Start":"01:30.170 ","End":"01:34.730","Text":"There\u0027s a fractional abundances 0.5069,"},{"Start":"01:34.730 ","End":"01:37.130","Text":"plus the mass of the second isotope,"},{"Start":"01:37.130 ","End":"01:38.570","Text":"which is actually what we\u0027re looking for,"},{"Start":"01:38.570 ","End":"01:45.590","Text":"the mass of bromine 81 times the fractional abundance of isotope 2,"},{"Start":"01:45.590 ","End":"01:47.750","Text":"which is 49.31 percent."},{"Start":"01:47.750 ","End":"01:51.430","Text":"Therefore the fractional abundance is 0.4931."},{"Start":"01:51.430 ","End":"01:59.150","Text":"Again, 79.904u equals the first contribution from the first isotope comes to 40u."},{"Start":"01:59.150 ","End":"02:02.360","Text":"Now we have to add the second contribution,"},{"Start":"02:02.360 ","End":"02:08.150","Text":"which is the mass of bromine 81, again times 0.4931."},{"Start":"02:08.150 ","End":"02:16.020","Text":"39.904u, the mass of bromine 81, times 0.4931."},{"Start":"02:16.020 ","End":"02:20.665","Text":"The mass of bromine 81 goes to 80.92u."},{"Start":"02:20.665 ","End":"02:22.835","Text":"That is our final answer."},{"Start":"02:22.835 ","End":"02:25.560","Text":"Thank you very much for listening."}],"ID":25214},{"Watched":false,"Name":"Exercise 6","Duration":"4m 12s","ChapterTopicVideoID":24392,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24392.jpeg","UploadDate":"2021-03-13T03:25:13.2570000","DurationForVideoObject":"PT4M12S","Description":null,"VideoComments":[],"Subtitles":[],"ID":25218},{"Watched":false,"Name":"Exercise 7","Duration":"3m 59s","ChapterTopicVideoID":24393,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24393.jpeg","UploadDate":"2021-03-13T03:25:24.6000000","DurationForVideoObject":"PT3M59S","Description":null,"VideoComments":[],"Subtitles":[],"ID":25219},{"Watched":false,"Name":"Exercise 8","Duration":"7m 8s","ChapterTopicVideoID":24391,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24391.jpeg","UploadDate":"2021-03-13T03:25:00.5330000","DurationForVideoObject":"PT7M8S","Description":null,"VideoComments":[],"Subtitles":[],"ID":25217},{"Watched":false,"Name":"Exercise 9","Duration":"4m 43s","ChapterTopicVideoID":24389,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24389.jpeg","UploadDate":"2021-03-13T03:24:33.1000000","DurationForVideoObject":"PT4M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:03.473","Text":"Hi. We are going to answer the following question"},{"Start":"00:03.473 ","End":"00:07.035","Text":": silver has 2 naturally occurring isotopes."},{"Start":"00:07.035 ","End":"00:12.060","Text":"Silver 107 has the mass of 106.905092u,"},{"Start":"00:12.060 ","End":"00:15.795","Text":"and a natural abundance of 51.84 percent."},{"Start":"00:15.795 ","End":"00:21.120","Text":"Calculate the mass and natural abundance of the other isotope, silver 109."},{"Start":"00:21.120 ","End":"00:24.645","Text":"In order to calculate the natural abundance of silver 109,"},{"Start":"00:24.645 ","End":"00:27.290","Text":"we will remember that the sum of"},{"Start":"00:27.290 ","End":"00:30.860","Text":"the natural abundances of all the isotopes equal 100 percent."},{"Start":"00:30.860 ","End":"00:37.190","Text":"Therefore, the natural abundance of silver 109 equals the total natural abundance,"},{"Start":"00:37.190 ","End":"00:38.869","Text":"which equals 100 percent,"},{"Start":"00:38.869 ","End":"00:43.460","Text":"minus the natural abundance of silver 107,"},{"Start":"00:43.460 ","End":"00:45.410","Text":"which is 51.84 percent,"},{"Start":"00:45.410 ","End":"00:49.260","Text":"and this equals 48.16 percent."},{"Start":"00:49.300 ","End":"00:53.345","Text":"That is the natural abundance of silver 109."},{"Start":"00:53.345 ","End":"00:56.105","Text":"Now, we will calculate the mass of silver 109."},{"Start":"00:56.105 ","End":"00:57.650","Text":"In order to calculate its mass,"},{"Start":"00:57.650 ","End":"01:03.395","Text":"we will use the formula atomic mass equals the mass of the first isotope,"},{"Start":"01:03.395 ","End":"01:07.220","Text":"times the fractional abundance of the first isotope,"},{"Start":"01:07.220 ","End":"01:10.435","Text":"plus the mass of the second isotope,"},{"Start":"01:10.435 ","End":"01:16.030","Text":"times the fractional abundance of the second isotope, and so on."},{"Start":"01:16.030 ","End":"01:18.190","Text":"In our case, since we only have 2 isotopes,"},{"Start":"01:18.190 ","End":"01:22.690","Text":"we will have the contribution for the first isotope and the second isotope."},{"Start":"01:22.690 ","End":"01:26.755","Text":"First of all, the atomic mass of silver is found in the periodic table of"},{"Start":"01:26.755 ","End":"01:32.000","Text":"elements and equals 107.868u."},{"Start":"01:32.000 ","End":"01:36.190","Text":"This equals the mass of the first isotope which is given and equals"},{"Start":"01:36.190 ","End":"01:42.010","Text":"106.905092u times the fractional abundance of the first isotope, which has given."},{"Start":"01:42.010 ","End":"01:45.220","Text":"The natural abundance equals 51.84 percent,"},{"Start":"01:45.220 ","End":"01:51.860","Text":"so the fractional abundance is 0.5184 plus the mass of the second isotope,"},{"Start":"01:51.860 ","End":"01:57.665","Text":"which is the mass of silver 109 times the fractional abundance of silver 109."},{"Start":"01:57.665 ","End":"02:01.248","Text":"We said that the natural abundance of silver 109 is 48.16"},{"Start":"02:01.248 ","End":"02:05.735","Text":"percent and therefore the fractional abundance equals 0.4816."},{"Start":"02:05.735 ","End":"02:14.360","Text":"Again, 107.868u equals, this comes to 55.42u plus,"},{"Start":"02:14.360 ","End":"02:21.960","Text":"again, the mass of silver 109 times 0.4816."},{"Start":"02:23.350 ","End":"02:31.560","Text":"So 52.448u equals the mass"},{"Start":"02:31.560 ","End":"02:36.610","Text":"of silver 109 times 0.4816."},{"Start":"02:37.550 ","End":"02:45.525","Text":"The mass of silver 109, comes to 108.9u."},{"Start":"02:45.525 ","End":"02:52.837","Text":"The natural abundance of silver 109 equals 48.16 percent,"},{"Start":"02:52.837 ","End":"02:57.300","Text":"and the mass of silver 109 equals 108.9u."},{"Start":"02:57.300 ","End":"02:59.000","Text":"Those are our final answers."},{"Start":"02:59.000 ","End":"03:01.560","Text":"Thank you very much for listening."}],"ID":25215},{"Watched":false,"Name":"Exercise 10","Duration":"2m 3s","ChapterTopicVideoID":24390,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24390.jpeg","UploadDate":"2021-03-13T03:24:36.9170000","DurationForVideoObject":"PT2M3S","Description":null,"VideoComments":[],"Subtitles":[],"ID":25216},{"Watched":false,"Name":"Exercise 11","Duration":"1m 35s","ChapterTopicVideoID":22837,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22837.jpeg","UploadDate":"2020-12-10T04:59:03.6570000","DurationForVideoObject":"PT1M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.315","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.315 ","End":"00:09.554","Text":"Identify the isotope that has 7 more neutrons than protons and a mass number of 59."},{"Start":"00:09.554 ","End":"00:13.950","Text":"We know that our isotope has a mass number of 59 meaning A,"},{"Start":"00:13.950 ","End":"00:18.945","Text":"which equals the number of neutrons plus protons, equals 59."},{"Start":"00:18.945 ","End":"00:23.750","Text":"We also know that there are 7 more neutrons than protons,"},{"Start":"00:23.750 ","End":"00:28.895","Text":"meaning the number of neutrons equals the number of protons plus 7."},{"Start":"00:28.895 ","End":"00:31.830","Text":"Now we\u0027re going to take our first equation,"},{"Start":"00:32.200 ","End":"00:37.025","Text":"A=n plus p. Instead of writing n,"},{"Start":"00:37.025 ","End":"00:41.130","Text":"we\u0027re going to write p plus 7 because we know that n=p plus 7."},{"Start":"00:42.500 ","End":"00:51.600","Text":"P plus 7 plus p"},{"Start":"00:51.600 ","End":"00:52.880","Text":"and this equals 2p plus 7 and this equals 59 because we know that"},{"Start":"00:52.880 ","End":"00:55.250","Text":"the mass number equals 59."},{"Start":"00:55.250 ","End":"00:58.940","Text":"Again, we have 2p plus 7=59."},{"Start":"00:58.940 ","End":"01:04.650","Text":"After solving for p, p=26."},{"Start":"01:04.650 ","End":"01:08.060","Text":"We know that the number of protons equals 26."},{"Start":"01:08.060 ","End":"01:11.585","Text":"The number of protons is the atomic number which equals z."},{"Start":"01:11.585 ","End":"01:15.230","Text":"We know that the atomic number of our isotope equals 26."},{"Start":"01:15.230 ","End":"01:16.820","Text":"If we look at the periodic table,"},{"Start":"01:16.820 ","End":"01:21.215","Text":"you will find that the atomic number 26 belongs to iron, Fe."},{"Start":"01:21.215 ","End":"01:25.020","Text":"The element is iron and the mass number is 59."},{"Start":"01:25.020 ","End":"01:27.665","Text":"We\u0027re going to write that in the upper-left corner."},{"Start":"01:27.665 ","End":"01:31.995","Text":"That\u0027s the isotope, iron 59."},{"Start":"01:31.995 ","End":"01:33.515","Text":"That\u0027s our final answer."},{"Start":"01:33.515 ","End":"01:36.150","Text":"Thank you very much for watching."}],"ID":23642},{"Watched":false,"Name":"Exercise 12","Duration":"2m 19s","ChapterTopicVideoID":24386,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/24386.jpeg","UploadDate":"2021-03-13T03:24:02.3330000","DurationForVideoObject":"PT2M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.550 ","End":"00:06.135","Text":"The following data were obtained for compounds of sulfur and oxygen."},{"Start":"00:06.135 ","End":"00:11.205","Text":"We have compound A with 2 grams of sulfur and 1.996 grams of oxygen."},{"Start":"00:11.205 ","End":"00:16.650","Text":"We have compound B with half a gram of sulfur and 0.7485 grams of oxygen."},{"Start":"00:16.650 ","End":"00:21.420","Text":"A, show that these results are consistent with Dalton\u0027s law of multiple proportions."},{"Start":"00:21.420 ","End":"00:24.420","Text":"B, if the formula of compound A is SO_2,"},{"Start":"00:24.420 ","End":"00:26.940","Text":"what is the formula of compound B?"},{"Start":"00:26.940 ","End":"00:30.000","Text":"We\u0027re going to start with A and we want to see that"},{"Start":"00:30.000 ","End":"00:34.125","Text":"our results are consistent with Dalton\u0027s Law of Multiple Proportions."},{"Start":"00:34.125 ","End":"00:38.385","Text":"So Dalton\u0027s Law of Multiple Proportions states that if 2 elements,"},{"Start":"00:38.385 ","End":"00:40.365","Text":"here we have sulfur and oxygen,"},{"Start":"00:40.365 ","End":"00:42.020","Text":"form more than a single compounds."},{"Start":"00:42.020 ","End":"00:43.355","Text":"So here they form A and B,"},{"Start":"00:43.355 ","End":"00:44.645","Text":"they form 2 compounds."},{"Start":"00:44.645 ","End":"00:47.570","Text":"The masses of one element combined with"},{"Start":"00:47.570 ","End":"00:52.300","Text":"a fixed mass of the second are in the ratio of small whole numbers."},{"Start":"00:52.300 ","End":"00:56.810","Text":"We have compound A and as we said,"},{"Start":"00:56.810 ","End":"00:59.000","Text":"we have sulfur and we have oxygen."},{"Start":"00:59.000 ","End":"01:04.610","Text":"We have 2 grams of sulfur and 1.996 grams of oxygen as written here."},{"Start":"01:04.610 ","End":"01:07.650","Text":"Now we\u0027re going to take a fixed mass of the sulfur,"},{"Start":"01:07.650 ","End":"01:11.925","Text":"meaning in compound B we\u0027re also going to take 2 grams."},{"Start":"01:11.925 ","End":"01:14.240","Text":"In order to get 2 grams and compound B,"},{"Start":"01:14.240 ","End":"01:17.065","Text":"we have to multiply our half a gram by 4."},{"Start":"01:17.065 ","End":"01:22.245","Text":"We have to multiply the mass of the oxygen by 4 also."},{"Start":"01:22.245 ","End":"01:27.890","Text":"We get 2 grams of sulfur and then the oxygen we have, let\u0027s write it down here,"},{"Start":"01:27.890 ","End":"01:35.940","Text":"0.7485 grams times 4 equals 2.994 grams of oxygen."},{"Start":"01:35.940 ","End":"01:40.580","Text":"As we see, we took a fixed mass of one element and now we have to"},{"Start":"01:40.580 ","End":"01:46.700","Text":"check if the ratio of the oxygen is in the ratio of small whole numbers."},{"Start":"01:46.700 ","End":"01:50.280","Text":"Meaning, let\u0027s take the ratio of oxygen."},{"Start":"01:50.710 ","End":"01:53.675","Text":"For example, between B and A,"},{"Start":"01:53.675 ","End":"02:01.260","Text":"it\u0027s 2.994 divided by 1.996."},{"Start":"02:01.260 ","End":"02:03.290","Text":"We can write the grams,"},{"Start":"02:03.290 ","End":"02:07.175","Text":"the grams cancel out, and this equals 1.5."},{"Start":"02:07.175 ","End":"02:08.870","Text":"If we put this in a fraction,"},{"Start":"02:08.870 ","End":"02:10.510","Text":"it equals 3-2,"},{"Start":"02:10.510 ","End":"02:15.125","Text":"meaning the oxygen is a ratio of small whole numbers, 3-2."},{"Start":"02:15.125 ","End":"02:19.430","Text":"Again, we have 2 elements and they form more than one compound."},{"Start":"02:19.430 ","End":"02:21.200","Text":"They form A and B."},{"Start":"02:21.200 ","End":"02:25.880","Text":"We took a fixed mass of one element and we\u0027re checking if the masses of"},{"Start":"02:25.880 ","End":"02:28.460","Text":"the other element of the oxygen are in the ratio"},{"Start":"02:28.460 ","End":"02:31.390","Text":"of small whole numbers and we found that they are."},{"Start":"02:31.390 ","End":"02:36.320","Text":"Therefore, our results are consistent with Dalton\u0027s Law of Multiple Proportions."},{"Start":"02:36.320 ","End":"02:38.415","Text":"Now we\u0027re going to go on to b."},{"Start":"02:38.415 ","End":"02:45.840","Text":"In b, they tell us that compound A is SO_2 and we\u0027re asked to find compound B."},{"Start":"02:45.840 ","End":"02:48.575","Text":"Again, when we took a fixed mass of the sulfur,"},{"Start":"02:48.575 ","End":"02:52.730","Text":"we can see that the ratio of the sulfur between B to"},{"Start":"02:52.730 ","End":"03:02.480","Text":"A equals 2 grams to 2 grams, which equals 1."},{"Start":"03:02.480 ","End":"03:06.770","Text":"So the ratio of the masses of the sulfur is 1 and the ratio of the masses of"},{"Start":"03:06.770 ","End":"03:12.855","Text":"the oxygen we already calculated between B to A is 3-2,"},{"Start":"03:12.855 ","End":"03:18.615","Text":"meaning the sulfur stays the same but the oxygen is"},{"Start":"03:18.615 ","End":"03:25.145","Text":"O_3 since the ratio of the oxygen is 3-2 and the ratio of the sulfur is 1-1."},{"Start":"03:25.145 ","End":"03:28.055","Text":"The compound in B is SO_3."},{"Start":"03:28.055 ","End":"03:29.525","Text":"That is our final answer."},{"Start":"03:29.525 ","End":"03:31.980","Text":"Thank you very much for watching."}],"ID":25212},{"Watched":false,"Name":"Exercise 13","Duration":"58s","ChapterTopicVideoID":22834,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22834.jpeg","UploadDate":"2020-12-10T04:55:34.6770000","DurationForVideoObject":"PT58S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23646},{"Watched":false,"Name":"Atomic Mass","Duration":"9m 1s","ChapterTopicVideoID":16903,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16903.jpeg","UploadDate":"2019-02-19T05:51:30.6400000","DurationForVideoObject":"PT9M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.430 ","End":"00:07.740","Text":"In this video, we will discuss the mass of a particular isotope or the atomic mass,"},{"Start":"00:07.740 ","End":"00:11.979","Text":"which takes into account the various isotopes of the element."},{"Start":"00:11.990 ","End":"00:15.915","Text":"Let\u0027s first consider the isotopic mass."},{"Start":"00:15.915 ","End":"00:20.415","Text":"That\u0027s the mass of a particular isotope of the element."},{"Start":"00:20.415 ","End":"00:24.300","Text":"The first thing to notice is that the actual mass of"},{"Start":"00:24.300 ","End":"00:29.580","Text":"an isotope is smaller than the total mass of all the particles."},{"Start":"00:29.580 ","End":"00:31.620","Text":"That means the mass of"},{"Start":"00:31.620 ","End":"00:40.050","Text":"the isotope is smaller"},{"Start":"00:40.050 ","End":"00:45.065","Text":"than the mass of all the neutrons,"},{"Start":"00:45.065 ","End":"00:49.315","Text":"protons, and electrons in the atom."},{"Start":"00:49.315 ","End":"00:52.555","Text":"Now, why might this be so?"},{"Start":"00:52.555 ","End":"01:00.800","Text":"The solution was found by Einstein using his famous equation, E equals mc^2."},{"Start":"01:00.800 ","End":"01:05.889","Text":"In this equation, E is energy and m is mass."},{"Start":"01:05.889 ","End":"01:10.820","Text":"Then when mass is lost, energy is produced."},{"Start":"01:10.820 ","End":"01:14.510","Text":"This is what happens when atoms combine in"},{"Start":"01:14.510 ","End":"01:19.640","Text":"the sun and we get a tremendous amount of energy emitted."},{"Start":"01:19.640 ","End":"01:24.380","Text":"Now how can we determine the mass of an isotope?"},{"Start":"01:24.380 ","End":"01:33.300","Text":"The isotopic mass is found experimentally using an instrument called a mass spectrometer."},{"Start":"01:34.130 ","End":"01:41.488","Text":"This is a very large and expensive instrument found in most chemistry departments,"},{"Start":"01:41.488 ","End":"01:46.505","Text":"and it\u0027s able to measure the mass of atoms and molecules."},{"Start":"01:46.505 ","End":"01:51.625","Text":"Now, there\u0027s an international body called IUPAC."},{"Start":"01:51.625 ","End":"01:57.935","Text":"That means the International Union of Pure and Applied Chemistry."},{"Start":"01:57.935 ","End":"02:04.315","Text":"They set rules in chemistry which are kept by all the chemists all over the world."},{"Start":"02:04.315 ","End":"02:11.690","Text":"One of the rules that\u0027s very important is that the mass of the isotope C12,"},{"Start":"02:11.690 ","End":"02:14.150","Text":"that\u0027s the one in which E,"},{"Start":"02:14.150 ","End":"02:16.384","Text":"the mass number is 12,"},{"Start":"02:16.384 ","End":"02:18.530","Text":"is defined as exactly,"},{"Start":"02:18.530 ","End":"02:22.520","Text":"precisely 12 atomic mass units,"},{"Start":"02:22.520 ","End":"02:25.175","Text":"which we write as 12u."},{"Start":"02:25.175 ","End":"02:27.770","Text":"Now this is by agreement."},{"Start":"02:27.770 ","End":"02:33.655","Text":"All other masses are relative to that of C12."},{"Start":"02:33.655 ","End":"02:38.675","Text":"Now before we consider the masses of some other atoms,"},{"Start":"02:38.675 ","End":"02:43.700","Text":"let\u0027s recall some few things about the masses of particles."},{"Start":"02:43.700 ","End":"02:52.725","Text":"Let\u0027s recall that the mass of a proton is very very close to 1 atomic mass unit,"},{"Start":"02:52.725 ","End":"02:59.735","Text":"and the mass of a neutron is also very very close to 1 atomic mass unit."},{"Start":"02:59.735 ","End":"03:05.195","Text":"Whereas the mass of an electron is very very small, almost negligible."},{"Start":"03:05.195 ","End":"03:09.290","Text":"If we take all the protons and count 1 for"},{"Start":"03:09.290 ","End":"03:13.625","Text":"each of them and all the neutrons and count 1 for each of them,"},{"Start":"03:13.625 ","End":"03:18.725","Text":"we will have the sum of all the neutrons and protons,"},{"Start":"03:18.725 ","End":"03:20.840","Text":"sum of all the neutrons,"},{"Start":"03:20.840 ","End":"03:24.085","Text":"number of neutrons plus the number of protons,"},{"Start":"03:24.085 ","End":"03:28.560","Text":"and recall that that is equal to A."},{"Start":"03:28.560 ","End":"03:36.350","Text":"We can conclude from this that the total mass of an isotope is very close to A,"},{"Start":"03:36.350 ","End":"03:38.120","Text":"which is the mass number,"},{"Start":"03:38.120 ","End":"03:41.345","Text":"that is the number of protons and neutrons."},{"Start":"03:41.345 ","End":"03:44.570","Text":"From considerations of Einstein,"},{"Start":"03:44.570 ","End":"03:50.530","Text":"we can expect it will be slightly smaller than A,"},{"Start":"03:50.530 ","End":"03:52.740","Text":"very very close to A,"},{"Start":"03:52.740 ","End":"03:55.960","Text":"but always slightly smaller than A."},{"Start":"03:55.960 ","End":"03:58.055","Text":"Let\u0027s take an example."},{"Start":"03:58.055 ","End":"04:02.780","Text":"The example we\u0027re going to consider is the isotope O16."},{"Start":"04:02.780 ","End":"04:06.575","Text":"This is the most common isotope of oxygen."},{"Start":"04:06.575 ","End":"04:15.530","Text":"Now the mass of O16 as measured in the mass spectrometer is 1.33291,"},{"Start":"04:15.530 ","End":"04:16.860","Text":"it\u0027s that accurate,"},{"Start":"04:16.860 ","End":"04:20.075","Text":"times the mass of the isotope C12."},{"Start":"04:20.075 ","End":"04:21.740","Text":"If we work that out,"},{"Start":"04:21.740 ","End":"04:25.565","Text":"it\u0027s 1.33291 times 12u."},{"Start":"04:25.565 ","End":"04:30.005","Text":"Remember 12u is the mass of the isotope C12."},{"Start":"04:30.005 ","End":"04:34.525","Text":"That comes out as 15.9949u,"},{"Start":"04:34.525 ","End":"04:37.455","Text":"which is approximately 16,"},{"Start":"04:37.455 ","End":"04:41.910","Text":"which is the mass number of oxygen."},{"Start":"04:41.910 ","End":"04:46.655","Text":"We see O16 has a mass number of 16,"},{"Start":"04:46.655 ","End":"04:50.675","Text":"but it\u0027s a tiny little bit smaller than that."},{"Start":"04:50.675 ","End":"04:54.530","Text":"That\u0027s in accordance with what Einstein said."},{"Start":"04:54.530 ","End":"04:57.844","Text":"Now we\u0027re going to consider the atomic mass."},{"Start":"04:57.844 ","End":"04:59.735","Text":"What\u0027s the atomic mass?"},{"Start":"04:59.735 ","End":"05:02.615","Text":"The atomic mass takes into account"},{"Start":"05:02.615 ","End":"05:07.745","Text":"all the isotopes of the element and their natural abundances."},{"Start":"05:07.745 ","End":"05:10.962","Text":"The atomic mass takes into account"},{"Start":"05:10.962 ","End":"05:16.910","Text":"the isotopes of an element and their natural abundances. What does this mean?"},{"Start":"05:16.910 ","End":"05:19.775","Text":"Let\u0027s take an example to show what we mean."},{"Start":"05:19.775 ","End":"05:26.135","Text":"An example is the atomic mass of naturally occurring oxygen."},{"Start":"05:26.135 ","End":"05:35.315","Text":"Now the abundance of the isotope O16 in naturally occurring oxygen is 99.76 percent."},{"Start":"05:35.315 ","End":"05:38.270","Text":"That\u0027s overwhelming all the others."},{"Start":"05:38.270 ","End":"05:40.115","Text":"It\u0027s in vast majority."},{"Start":"05:40.115 ","End":"05:46.220","Text":"However, there are some other isotopes present O18 is 0.2"},{"Start":"05:46.220 ","End":"05:53.320","Text":"percent and O17 is even less than that, 0.04 percent."},{"Start":"05:53.320 ","End":"05:58.415","Text":"We can calculate something called the atomic mass."},{"Start":"05:58.415 ","End":"06:04.010","Text":"The atomic mass takes account of the fact that it\u0027s not just O16,"},{"Start":"06:04.010 ","End":"06:09.575","Text":"but also some O18 and some O17."},{"Start":"06:09.575 ","End":"06:16.240","Text":"Now what we\u0027re going to calculate is what\u0027s called a weighted average."},{"Start":"06:19.910 ","End":"06:28.585","Text":"A weighted average takes into account the percentages of all the isotopes."},{"Start":"06:28.585 ","End":"06:35.060","Text":"We see that its vast majority is O16 with a little O17 and O18,"},{"Start":"06:35.060 ","End":"06:37.370","Text":"which are heavier than O16."},{"Start":"06:37.370 ","End":"06:45.460","Text":"So we expect that the atomic mass will be a little greater than that of O16."},{"Start":"06:45.460 ","End":"06:48.695","Text":"Let\u0027s consider how we do this."},{"Start":"06:48.695 ","End":"06:50.720","Text":"The mass of oxygen,"},{"Start":"06:50.720 ","End":"06:58.190","Text":"naturally occurring oxygen is 99.76 percent times its actual mass,"},{"Start":"06:58.190 ","End":"07:01.760","Text":"which we saw before was 15.9949."},{"Start":"07:01.760 ","End":"07:11.235","Text":"That\u0027s the mass of O16 and 0.2 percent times the mass of O18,"},{"Start":"07:11.235 ","End":"07:15.090","Text":"which is 17.9992 and"},{"Start":"07:15.090 ","End":"07:23.310","Text":"0.04 times the mass of O17, which is 16.9991."},{"Start":"07:23.310 ","End":"07:31.125","Text":"Now these masses of O18 and O17 are experimentally determined."},{"Start":"07:31.125 ","End":"07:33.530","Text":"It\u0027s not something that we just know."},{"Start":"07:33.530 ","End":"07:36.727","Text":"When we do this calculation,"},{"Start":"07:36.727 ","End":"07:41.090","Text":"we get a number which we have to divide by"},{"Start":"07:41.090 ","End":"07:47.600","Text":"a 100 percent because the sum of all these percentages,"},{"Start":"07:47.600 ","End":"07:49.370","Text":"the sum of all these abundances,"},{"Start":"07:49.370 ","End":"07:56.945","Text":"99.76 plus 0.2 plus 0.04 is 100 percent."},{"Start":"07:56.945 ","End":"08:05.015","Text":"We divide by 100 and we get out the answer, 50.9993u."},{"Start":"08:05.015 ","End":"08:12.020","Text":"Now this number is slightly bigger than the mass of 016,"},{"Start":"08:12.020 ","End":"08:15.795","Text":"as we suggested, but not enormously so."},{"Start":"08:15.795 ","End":"08:21.725","Text":"That takes into account that there\u0027s very little O18 and O17."},{"Start":"08:21.725 ","End":"08:25.970","Text":"The answer comes out a little more than that of O16,"},{"Start":"08:25.970 ","End":"08:28.625","Text":"but not a great deal more."},{"Start":"08:28.625 ","End":"08:33.200","Text":"We should mention that sometimes people"},{"Start":"08:33.200 ","End":"08:39.890","Text":"see that this is the atomic weight rather than the atomic mass."},{"Start":"08:39.890 ","End":"08:43.970","Text":"It\u0027s a mistake, but a very common mistake."},{"Start":"08:43.970 ","End":"08:49.760","Text":"This is also called the atomic weight of oxygen."},{"Start":"08:49.760 ","End":"08:55.895","Text":"In this video, we have discussed the isotopic mass and the atomic mass,"},{"Start":"08:55.895 ","End":"08:59.580","Text":"sometimes called the atomic weight."}],"ID":17650},{"Watched":false,"Name":"Exercise 14","Duration":"48s","ChapterTopicVideoID":22835,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22835.jpeg","UploadDate":"2020-12-10T04:55:36.1530000","DurationForVideoObject":"PT48S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23647},{"Watched":false,"Name":"Exercise 15","Duration":"2m 20s","ChapterTopicVideoID":22836,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22836.jpeg","UploadDate":"2020-12-10T04:55:40.1730000","DurationForVideoObject":"PT2M20S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23648},{"Watched":false,"Name":"Exercise 16","Duration":"2m 24s","ChapterTopicVideoID":22831,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22831.jpeg","UploadDate":"2020-12-10T04:55:23.5000000","DurationForVideoObject":"PT2M24S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23643},{"Watched":false,"Name":"Exercise 17","Duration":"3m ","ChapterTopicVideoID":22832,"CourseChapterTopicPlaylistID":86810,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22832.jpeg","UploadDate":"2020-12-10T04:55:29.3630000","DurationForVideoObject":"PT3M","Description":null,"VideoComments":[],"Subtitles":[],"ID":23644}],"Thumbnail":null,"ID":86810},{"Name":"Calculations Involving Moles","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Mole and Avogadro number","Duration":"13m 18s","ChapterTopicVideoID":22850,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22850.jpeg","UploadDate":"2020-12-13T17:27:58.0270000","DurationForVideoObject":"PT13M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.855","Text":"In this video, we will discuss the important concepts of the mole and Avogadro\u0027s number."},{"Start":"00:06.855 ","End":"00:11.265","Text":"Let\u0027s begin by recalling some facts we learned in previous videos."},{"Start":"00:11.265 ","End":"00:18.360","Text":"The first thing to remember is that the mole is the SI unit for amount of substance."},{"Start":"00:18.360 ","End":"00:22.110","Text":"The second thing we need to remember is the mass of a C^12 atom."},{"Start":"00:22.110 ","End":"00:28.185","Text":"A single C^12 of an atom is precisely 12 atomic mass units,"},{"Start":"00:28.185 ","End":"00:31.200","Text":"we right atomic mass units by little u."},{"Start":"00:31.200 ","End":"00:39.990","Text":"The third thing is the mass of the O^16 atom is 15.9949 atomic mass units,"},{"Start":"00:39.990 ","End":"00:44.740","Text":"which is almost equal to 16 atomic mass units."},{"Start":"00:44.740 ","End":"00:50.765","Text":"The fourth thing is that the atomic mass unit is something that\u0027s extremely small."},{"Start":"00:50.765 ","End":"00:57.845","Text":"It\u0027s only 1.66 times 10 to the power minus 27 kilograms."},{"Start":"00:57.845 ","End":"01:00.500","Text":"That means that an atom,"},{"Start":"01:00.500 ","End":"01:04.145","Text":"a single atom, is very small."},{"Start":"01:04.145 ","End":"01:09.800","Text":"In this video, we will connect all these statements into one consistent whole."},{"Start":"01:09.800 ","End":"01:12.140","Text":"First, we should note that the mass of"},{"Start":"01:12.140 ","End":"01:16.465","Text":"a single carbon or oxygen atom is extremely small."},{"Start":"01:16.465 ","End":"01:23.284","Text":"That we would need an enormous number of them to produce 1 gram of carbon or oxygen."},{"Start":"01:23.284 ","End":"01:25.910","Text":"In order to overcome this difficulty,"},{"Start":"01:25.910 ","End":"01:30.560","Text":"chemists use the mole concept and Avogadro\u0027s number."},{"Start":"01:30.560 ","End":"01:34.130","Text":"Let\u0027s begin by defining these concepts."},{"Start":"01:34.130 ","End":"01:37.925","Text":"We\u0027re going to define the mole and Avogadro\u0027s number."},{"Start":"01:37.925 ","End":"01:39.965","Text":"Here\u0027s the definition of the mole,"},{"Start":"01:39.965 ","End":"01:46.760","Text":"1 mole of C^12 atoms has a mass of precisely 12 grams."},{"Start":"01:46.760 ","End":"01:52.465","Text":"C^12 is used as a standard for such calculations."},{"Start":"01:52.465 ","End":"01:59.285","Text":"1 mole of C^12 atoms has a mass of precisely 12 grams."},{"Start":"01:59.285 ","End":"02:03.380","Text":"Now the number of particles in 1 mole of C^12,"},{"Start":"02:03.380 ","End":"02:06.110","Text":"that is in 12 grams,"},{"Start":"02:06.110 ","End":"02:08.705","Text":"is called Avogadro\u0027s number."},{"Start":"02:08.705 ","End":"02:13.130","Text":"It\u0027s a number with great which has been calculated degree position."},{"Start":"02:13.130 ","End":"02:16.160","Text":"Any Avogadro\u0027s number is"},{"Start":"02:16.160 ","End":"02:23.210","Text":"6.022140857 times"},{"Start":"02:23.210 ","End":"02:27.290","Text":"10^23 atoms per mole."},{"Start":"02:27.290 ","End":"02:36.195","Text":"In this course, we\u0027ll just write N_A is approximately 6.022 times 10^23."},{"Start":"02:36.195 ","End":"02:42.590","Text":"We can conclude from this that the mass of Avogadro number of C^12 atoms,"},{"Start":"02:42.590 ","End":"02:49.085","Text":"that is the mass of 1 mole is precisely 12 grams."},{"Start":"02:49.085 ","End":"02:55.160","Text":"Now Avogadro\u0027s number is a very large number and it\u0027s extremely difficult to imagine."},{"Start":"02:55.160 ","End":"02:58.295","Text":"In order to understand how we use it."},{"Start":"02:58.295 ","End":"03:04.825","Text":"Let\u0027s consider an analogy and the analogy we\u0027re going to use is the dozen."},{"Start":"03:04.825 ","End":"03:07.565","Text":"Dozen is a very simple number."},{"Start":"03:07.565 ","End":"03:11.075","Text":"We know that the dozen is just 12."},{"Start":"03:11.075 ","End":"03:13.790","Text":"It\u0027s like Avogadro\u0027s number,"},{"Start":"03:13.790 ","End":"03:15.950","Text":"and this is a fixed number,"},{"Start":"03:15.950 ","End":"03:20.555","Text":"but a very small number that\u0027s easy to deal with."},{"Start":"03:20.555 ","End":"03:26.885","Text":"Here we have a picture of an egg box containing 12 eggs."},{"Start":"03:26.885 ","End":"03:30.055","Text":"Now let\u0027s consider an egg,"},{"Start":"03:30.055 ","End":"03:34.714","Text":"a single egg, and imagine it has a mass of 60 grams."},{"Start":"03:34.714 ","End":"03:36.890","Text":"Then the mass of a dozen eggs,"},{"Start":"03:36.890 ","End":"03:40.625","Text":"we can calculate to be 12 times 60 grams,"},{"Start":"03:40.625 ","End":"03:43.430","Text":"which is 720 grams."},{"Start":"03:43.430 ","End":"03:47.120","Text":"Now, I want to compare eggs and apples."},{"Start":"03:47.120 ","End":"03:50.945","Text":"Apples are larger than eggs."},{"Start":"03:50.945 ","End":"03:55.805","Text":"Let\u0027s assume that an apple has a mass of 120 grams."},{"Start":"03:55.805 ","End":"04:00.590","Text":"Then the mass of a dozen apples is 12 times 120 grams,"},{"Start":"04:00.590 ","End":"04:08.515","Text":"which is 1440 grams and I want to compare the mass of an apple to the mass of an egg."},{"Start":"04:08.515 ","End":"04:15.095","Text":"The mass of 1 apple compared to the mass of 1 egg is 120 grams divided by 60 grams."},{"Start":"04:15.095 ","End":"04:21.515","Text":"We can cancel the grams,120 divided by 60 is just 2."},{"Start":"04:21.515 ","End":"04:24.110","Text":"Now let\u0027s take a dozen apples."},{"Start":"04:24.110 ","End":"04:28.325","Text":"We\u0027ll need rather large egg boxes or apple boxes."},{"Start":"04:28.325 ","End":"04:31.115","Text":"Compared to the mass of a dozen eggs."},{"Start":"04:31.115 ","End":"04:36.440","Text":"That\u0027s 12 times 120 grams divided by 12 times 60 grams."},{"Start":"04:36.440 ","End":"04:45.290","Text":"Now, the important thing is that the 12s cancel and the grams cancel."},{"Start":"04:45.290 ","End":"04:53.695","Text":"We\u0027re once again, we\u0027re left with 120 divided by 60 and we know that that\u0027s just 2."},{"Start":"04:53.695 ","End":"04:59.481","Text":"We see that it doesn\u0027t matter how many apples and eggs we have,"},{"Start":"04:59.481 ","End":"05:02.795","Text":"the ratio is just the same."},{"Start":"05:02.795 ","End":"05:04.905","Text":"Here it\u0027s 2,"},{"Start":"05:04.905 ","End":"05:06.935","Text":"and here it is again 2."},{"Start":"05:06.935 ","End":"05:11.290","Text":"Now let\u0027s go one step further and consider the mass of"},{"Start":"05:11.290 ","End":"05:18.145","Text":"Avogadro\u0027s numbers of apples compared to the mass of Avogadro\u0027s number of eggs,"},{"Start":"05:18.145 ","End":"05:20.680","Text":"the enormous number of apples and eggs."},{"Start":"05:20.680 ","End":"05:28.240","Text":"We can write the mass of Avogadro\u0027s number of apples as N_A times 120 grams,"},{"Start":"05:28.240 ","End":"05:33.100","Text":"and Avogadro\u0027s number of eggs as N_A times 60 grams."},{"Start":"05:33.100 ","End":"05:38.780","Text":"The N_A cancels, the grams cancel,"},{"Start":"05:38.780 ","End":"05:42.260","Text":"we\u0027re left again with a 120 over 60,"},{"Start":"05:42.260 ","End":"05:44.830","Text":"which is just equal to 2."},{"Start":"05:44.830 ","End":"05:51.500","Text":"That\u0027s we see that the ratio is 2 regardless of how many apples and eggs we have,"},{"Start":"05:51.500 ","End":"05:55.670","Text":"provided we have the same number of apples and eggs."},{"Start":"05:55.670 ","End":"05:57.920","Text":"Now how are we going to use all this?"},{"Start":"05:57.920 ","End":"06:00.830","Text":"Well, instead of comparing apples and eggs,"},{"Start":"06:00.830 ","End":"06:07.115","Text":"now we\u0027re going to compare two different sorts of atoms."},{"Start":"06:07.115 ","End":"06:12.110","Text":"We\u0027re going to compare them so we can find the masses of other atoms."},{"Start":"06:12.110 ","End":"06:15.715","Text":"That\u0027s atoms that are not C^12."},{"Start":"06:15.715 ","End":"06:19.540","Text":"We\u0027re going to take as an example the mass of"},{"Start":"06:19.540 ","End":"06:25.360","Text":"O^16 atom and compare it to the mass of a C^12 atom."},{"Start":"06:25.360 ","End":"06:29.195","Text":"Once again, let\u0027s write the ratio,"},{"Start":"06:29.195 ","End":"06:34.760","Text":"the mass of an O^16 atom compared to the mass of a C^12 atom."},{"Start":"06:34.760 ","End":"06:41.390","Text":"We learned before the mass of an O^16 atom is almost exactly 16 atomic mass units,"},{"Start":"06:41.390 ","End":"06:48.095","Text":"and the mass of the C^12 atom is precisely 12 atomic mass units."},{"Start":"06:48.095 ","End":"06:53.300","Text":"The atomic mass units go and we\u0027re left with 16 divided by 12,"},{"Start":"06:53.300 ","End":"06:56.315","Text":"which of course is 4 divided by 3."},{"Start":"06:56.315 ","End":"07:01.310","Text":"Now let\u0027s go one step further and take the mass of Avogadro\u0027s number of"},{"Start":"07:01.310 ","End":"07:07.804","Text":"O^16 atoms compare it to the mass of Avogadro\u0027s number of C^12 atoms."},{"Start":"07:07.804 ","End":"07:12.425","Text":"The mass of Avogadro\u0027s number of O^16 atoms is"},{"Start":"07:12.425 ","End":"07:17.570","Text":"N_A times 16 atomic mass units and of C^12 atoms,"},{"Start":"07:17.570 ","End":"07:22.945","Text":"it\u0027s N_A times 12 atomic mass units."},{"Start":"07:22.945 ","End":"07:27.265","Text":"The N_A\u0027s go and U\u0027s go,"},{"Start":"07:27.265 ","End":"07:29.525","Text":"we\u0027re left with 16 over 12,"},{"Start":"07:29.525 ","End":"07:33.760","Text":"which once again is just 4 over 3."},{"Start":"07:33.760 ","End":"07:37.930","Text":"We see that it doesn\u0027t matter how many atoms we have,"},{"Start":"07:37.930 ","End":"07:40.120","Text":"the ratio will always be the same."},{"Start":"07:40.120 ","End":"07:45.190","Text":"The ratio will be 4 over 3 here,"},{"Start":"07:45.190 ","End":"07:47.155","Text":"4 over 3 here."},{"Start":"07:47.155 ","End":"07:55.755","Text":"Now the mass of Avogadro\u0027s number of C^12 atoms is just the mass of a mole of C^12 atoms."},{"Start":"07:55.755 ","End":"07:57.745","Text":"We know by definition,"},{"Start":"07:57.745 ","End":"08:02.680","Text":"the mass of a mole of C^12 atoms is 12 grams."},{"Start":"08:02.680 ","End":"08:09.070","Text":"Now, what about O^16?"},{"Start":"08:09.070 ","End":"08:13.450","Text":"The mass of Avogadro\u0027s number of O^16 atoms is the same as"},{"Start":"08:13.450 ","End":"08:18.485","Text":"the mass of a mole of O^16 atoms and that\u0027s what we want to find out."},{"Start":"08:18.485 ","End":"08:21.710","Text":"We\u0027re going to write that as x grams."},{"Start":"08:21.710 ","End":"08:24.755","Text":"That\u0027s what we\u0027re going to determine."},{"Start":"08:24.755 ","End":"08:27.469","Text":"Once again, we have a ratio."},{"Start":"08:27.469 ","End":"08:29.675","Text":"Now, mass of N_A,"},{"Start":"08:29.675 ","End":"08:36.200","Text":"O^16 atoms compared to the mass of N_A C^12 atoms is the"},{"Start":"08:36.200 ","End":"08:43.420","Text":"same as the mass of moles of O^16 atoms compared to the mass of a mole of C^12 atoms."},{"Start":"08:43.420 ","End":"08:48.610","Text":"That\u0027s x grams because we said the mass of a mole of O^16 atoms would"},{"Start":"08:48.610 ","End":"08:54.340","Text":"be x grams and we know that the mass of a mole of C^12 atoms is 12 grams."},{"Start":"08:54.340 ","End":"08:58.610","Text":"We can cancel the grams and we\u0027re left x over"},{"Start":"08:58.610 ","End":"09:04.670","Text":"12 and we know from what we had before that this ratio,"},{"Start":"09:04.670 ","End":"09:08.150","Text":"the ratio of the mass of Avogadro\u0027s number of"},{"Start":"09:08.150 ","End":"09:13.355","Text":"O^16 atoms compared to the mass of Avogadro\u0027s numbers of C^12 atoms,"},{"Start":"09:13.355 ","End":"09:17.580","Text":"we know that that is equal to 4 over 3."},{"Start":"09:17.740 ","End":"09:25.130","Text":"Now we have x over 12 is equal to 4 over 3,"},{"Start":"09:25.130 ","End":"09:31.250","Text":"which means that x is equal to 48 divided by"},{"Start":"09:31.250 ","End":"09:40.135","Text":"3 which is just 16."},{"Start":"09:40.135 ","End":"09:43.595","Text":"What can we conclude from all this?"},{"Start":"09:43.595 ","End":"09:49.650","Text":"We can conclude that the molar mass"},{"Start":"09:51.490 ","End":"10:02.489","Text":"of oxygen 16 is 16 grams and previously we learned that the atomic mass,"},{"Start":"10:02.680 ","End":"10:13.740","Text":"the mass of 1 atom of O^16 is 16 atomic mass units, 16u."},{"Start":"10:13.740 ","End":"10:15.925","Text":"So this is very interesting."},{"Start":"10:15.925 ","End":"10:20.245","Text":"We see that it\u0027s precisely the same number."},{"Start":"10:20.245 ","End":"10:23.950","Text":"The molar mass is 16 grams,"},{"Start":"10:23.950 ","End":"10:27.550","Text":"and the atomic mass is 16 atomic mass units."},{"Start":"10:27.550 ","End":"10:30.400","Text":"From this, we can make a rule."},{"Start":"10:30.400 ","End":"10:33.820","Text":"The rule is that if we look at tables of"},{"Start":"10:33.820 ","End":"10:38.755","Text":"atomic masses that appear in textbooks or in the Internet,"},{"Start":"10:38.755 ","End":"10:41.290","Text":"we can read them in two ways."},{"Start":"10:41.290 ","End":"10:44.125","Text":"The same number can be read in two ways."},{"Start":"10:44.125 ","End":"10:51.280","Text":"It\u0027s either the atomic mass in atomic mass units or it\u0027s the molar mass in grams."},{"Start":"10:51.280 ","End":"10:55.160","Text":"This is very important and we\u0027ll use it all the time."},{"Start":"10:55.160 ","End":"10:58.040","Text":"Now for the last part of this video,"},{"Start":"10:58.040 ","End":"11:01.430","Text":"we\u0027re going to calculate the atomic mass unit."},{"Start":"11:01.430 ","End":"11:06.170","Text":"We\u0027re going to use what we learned to calculate the atomic mass unit."},{"Start":"11:06.170 ","End":"11:12.410","Text":"Now we know the mass of a single C^12 atom is going to"},{"Start":"11:12.410 ","End":"11:18.710","Text":"be the mass of Avogadro\u0027s number of C^12 atoms,"},{"Start":"11:18.710 ","End":"11:20.750","Text":"which is 12 grams,"},{"Start":"11:20.750 ","End":"11:23.350","Text":"divided by Avogadro\u0027s number."},{"Start":"11:23.350 ","End":"11:31.550","Text":"We\u0027re going to divide 12 grams by 6.22 times 10^23 atoms."},{"Start":"11:31.550 ","End":"11:34.025","Text":"We know that by definition,"},{"Start":"11:34.025 ","End":"11:38.885","Text":"this must be equal to 12 atomic mass units."},{"Start":"11:38.885 ","End":"11:41.720","Text":"Now, we can get rid of the 12s."},{"Start":"11:41.720 ","End":"11:44.330","Text":"We can cancel 12 here."},{"Start":"11:44.330 ","End":"11:48.095","Text":"Divide both sides of the equation by 12,"},{"Start":"11:48.095 ","End":"11:55.705","Text":"divide 1 gram by 6.022 times 10^23."},{"Start":"11:55.705 ","End":"12:00.690","Text":"That gives us 1 atomic mass unit and the division is"},{"Start":"12:00.690 ","End":"12:07.665","Text":"1.66 times 10 to the power minus 24 and the grams."},{"Start":"12:07.665 ","End":"12:10.205","Text":"This is the atomic mass unit."},{"Start":"12:10.205 ","End":"12:18.200","Text":"1 atomic mass unit is 1.66 times 10 to the minus 24 grams."},{"Start":"12:18.200 ","End":"12:22.550","Text":"Usually, it\u0027s written in terms of kilograms."},{"Start":"12:22.550 ","End":"12:26.480","Text":"We have to turn these grams into kilograms."},{"Start":"12:26.480 ","End":"12:28.505","Text":"How do we do that?"},{"Start":"12:28.505 ","End":"12:32.996","Text":"We write equal to 1.66 times 10 to the power"},{"Start":"12:32.996 ","End":"12:37.715","Text":"minus 24 grams and we use a conversion factor."},{"Start":"12:37.715 ","End":"12:43.110","Text":"The one that release between kilograms and grams,"},{"Start":"12:43.110 ","End":"12:47.595","Text":"1 kilogram is 10^3 grams."},{"Start":"12:47.595 ","End":"12:56.075","Text":"We have 1 kilogram divided by 10^3 grams and when we multiply out,"},{"Start":"12:56.075 ","End":"13:03.525","Text":"we get 1.66 times 10 to the minus 27 kilograms,"},{"Start":"13:03.525 ","End":"13:07.595","Text":"the grams cancel, we\u0027re left with kilograms."},{"Start":"13:07.595 ","End":"13:12.890","Text":"In this video, we discussed the definitions of the mole and Avogadro\u0027s number."},{"Start":"13:12.890 ","End":"13:17.610","Text":"We also calculated the atomic mass unit."}],"ID":23670},{"Watched":false,"Name":"Conversion Factors Involving Moles","Duration":"7m 14s","ChapterTopicVideoID":16902,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16902.jpeg","UploadDate":"2019-03-18T23:40:24.0730000","DurationForVideoObject":"PT7M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.270","Text":"In this video, we will show how mole, molar mass,"},{"Start":"00:04.270 ","End":"00:10.360","Text":"and Avogadro\u0027s number can be used as conversion factors that help us to solve problems."},{"Start":"00:10.360 ","End":"00:13.900","Text":"Let\u0027s recall some things that we\u0027ve learned before."},{"Start":"00:13.900 ","End":"00:18.130","Text":"The first thing we need to recall is that the mass of 1 mole,"},{"Start":"00:18.130 ","End":"00:19.615","Text":"that\u0027s the molar mass,"},{"Start":"00:19.615 ","End":"00:24.714","Text":"of C12 atoms is precisely 12 grams."},{"Start":"00:24.714 ","End":"00:28.960","Text":"The second thing we need to remember is that the mass of 1 mole,"},{"Start":"00:28.960 ","End":"00:30.385","Text":"that\u0027s the molar mass,"},{"Start":"00:30.385 ","End":"00:36.295","Text":"of O16 atoms is approximately 16 grams."},{"Start":"00:36.295 ","End":"00:37.900","Text":"That\u0027s approximately."},{"Start":"00:37.900 ","End":"00:40.090","Text":"Remember that for O16,"},{"Start":"00:40.090 ","End":"00:42.490","Text":"it\u0027s approximately 16 grams,"},{"Start":"00:42.490 ","End":"00:43.985","Text":"whereas for C12,"},{"Start":"00:43.985 ","End":"00:46.610","Text":"it\u0027s precisely 12 grams."},{"Start":"00:46.610 ","End":"00:49.820","Text":"The third thing we need to remember is that 1 mole of"},{"Start":"00:49.820 ","End":"00:54.320","Text":"any element contains Avogadro\u0027s number of atoms."},{"Start":"00:54.320 ","End":"01:04.385","Text":"We\u0027ll take Avogadro\u0027s number as approximately 6.022 times 10^23."},{"Start":"01:04.385 ","End":"01:07.955","Text":"Now, the first conversion factor we\u0027re going to consider"},{"Start":"01:07.955 ","End":"01:12.184","Text":"is that that relates mole to molar mass."},{"Start":"01:12.184 ","End":"01:14.405","Text":"We\u0027re going to consider an example."},{"Start":"01:14.405 ","End":"01:17.600","Text":"The example we\u0027re going to consider is carbon."},{"Start":"01:17.600 ","End":"01:20.509","Text":"We\u0027re going to ask a very simple question."},{"Start":"01:20.509 ","End":"01:24.430","Text":"What is the mass of 10 moles of C12?"},{"Start":"01:24.430 ","End":"01:26.570","Text":"Now, we need to recall,"},{"Start":"01:26.570 ","End":"01:29.000","Text":"which we said just a few minutes ago,"},{"Start":"01:29.000 ","End":"01:33.530","Text":"that the mass of 1 mole of carbon-12 has"},{"Start":"01:33.530 ","End":"01:38.790","Text":"a mass of"},{"Start":"01:38.790 ","End":"01:44.720","Text":"precisely 12 grams."},{"Start":"01:44.720 ","End":"01:46.880","Text":"Now, here\u0027s our conversion factor,"},{"Start":"01:46.880 ","End":"01:51.530","Text":"12 grams, which is equivalent to 1 mole of C12,"},{"Start":"01:51.530 ","End":"01:58.760","Text":"we can write as 12 grams divided by 1 mole of C12 equal to 1 or the inverse,"},{"Start":"01:58.760 ","End":"02:05.375","Text":"we can turn this upside down and write 1 mole of C12 divided by 12 grams equals to 1."},{"Start":"02:05.375 ","End":"02:08.855","Text":"Now we\u0027re in a position to solve the problem."},{"Start":"02:08.855 ","End":"02:14.010","Text":"The mass of 10 moles of C12 is 10 moles of"},{"Start":"02:14.010 ","End":"02:20.720","Text":"C12 times 12 grams divide by 1 mole of C12."},{"Start":"02:20.720 ","End":"02:24.980","Text":"Now, we\u0027ve chosen to write the conversion factor like this,"},{"Start":"02:24.980 ","End":"02:28.030","Text":"12 grams divide by 1 mole of C12,"},{"Start":"02:28.030 ","End":"02:36.830","Text":"because we want to be left at the end with a mass and to get rid of the number of moles."},{"Start":"02:36.830 ","End":"02:38.720","Text":"Here we have moles,"},{"Start":"02:38.720 ","End":"02:42.650","Text":"we can cancel, and we\u0027re left just with grams."},{"Start":"02:42.650 ","End":"02:48.140","Text":"Now we can multiply out 10 times 12, gives us 120."},{"Start":"02:48.140 ","End":"02:50.780","Text":"We\u0027re left just with grams."},{"Start":"02:50.780 ","End":"02:53.270","Text":"Here\u0027s our answer."},{"Start":"02:53.270 ","End":"02:55.940","Text":"Let\u0027s consider another example."},{"Start":"02:55.940 ","End":"03:00.955","Text":"This time the example is based on oxygen rather than carbon."},{"Start":"03:00.955 ","End":"03:03.335","Text":"It\u0027s just the same idea."},{"Start":"03:03.335 ","End":"03:05.315","Text":"Question is, once again,"},{"Start":"03:05.315 ","End":"03:09.050","Text":"what is the mass of 10 moles of O16?"},{"Start":"03:09.050 ","End":"03:13.700","Text":"Now, recall from just a few minutes ago that the mass of"},{"Start":"03:13.700 ","End":"03:18.965","Text":"1 mole of O16 is approximately 16 grams."},{"Start":"03:18.965 ","End":"03:20.900","Text":"Now we can solve the problem."},{"Start":"03:20.900 ","End":"03:25.600","Text":"The mass of 10 moles of O16 is 10 moles of O16."},{"Start":"03:25.600 ","End":"03:31.100","Text":"We really need to say moles of what times 16 grams divided"},{"Start":"03:31.100 ","End":"03:36.620","Text":"by 1 mole of O16 because 1 mole of O16 weighs 16 grams,"},{"Start":"03:36.620 ","End":"03:39.095","Text":"has a mass of 16 grams."},{"Start":"03:39.095 ","End":"03:42.865","Text":"Once again, we can cancel the moles."},{"Start":"03:42.865 ","End":"03:53.220","Text":"We\u0027re left just with grams and we can multiply 10 times 16 is 160 grams."},{"Start":"03:53.220 ","End":"03:56.405","Text":"Our answer is 160 grams."},{"Start":"03:56.405 ","End":"04:00.020","Text":"The next conversion factor we\u0027re going to talk about is"},{"Start":"04:00.020 ","End":"04:03.710","Text":"that that relates mole to Avogadro\u0027s number."},{"Start":"04:03.710 ","End":"04:07.250","Text":"The example we\u0027re going to take is carbon or"},{"Start":"04:07.250 ","End":"04:12.840","Text":"indeed any other element because Avogadro\u0027s number is the same for any element."},{"Start":"04:12.840 ","End":"04:18.335","Text":"The question is, how many atoms are there in 10 moles of C12,"},{"Start":"04:18.335 ","End":"04:20.930","Text":"or indeed any other element?"},{"Start":"04:20.930 ","End":"04:25.500","Text":"We need to recall that 1 mole"},{"Start":"04:25.570 ","End":"04:28.710","Text":"contains"},{"Start":"04:29.980 ","End":"04:37.830","Text":"6.022 times 10^23 atoms."},{"Start":"04:37.990 ","End":"04:41.165","Text":"Now we can write the conversion factor."},{"Start":"04:41.165 ","End":"04:45.860","Text":"1 mole divided by 6.022 times"},{"Start":"04:45.860 ","End":"04:55.190","Text":"10^23 atoms is equal to1 because 1 mole is equivalent to 6.022 times 10^23 atoms."},{"Start":"04:55.190 ","End":"04:57.364","Text":"Or we can write the inverse,"},{"Start":"04:57.364 ","End":"05:04.920","Text":"6.022 times 10^23 atoms divided by 1 mole equals 1."},{"Start":"05:04.960 ","End":"05:07.885","Text":"Now we can solve the problem."},{"Start":"05:07.885 ","End":"05:12.600","Text":"The number of atoms in 10 moles is equal to10 moles times"},{"Start":"05:12.600 ","End":"05:19.425","Text":"6.022 times 10^23 atoms divided by 1 mole,"},{"Start":"05:19.425 ","End":"05:24.530","Text":"which is 6.022 times 10^24 atoms."},{"Start":"05:24.530 ","End":"05:29.555","Text":"Once again, the moles cancel and we\u0027re left just with the atoms."},{"Start":"05:29.555 ","End":"05:37.175","Text":"Our answer is 6.022 times 10^24 atoms."},{"Start":"05:37.175 ","End":"05:43.800","Text":"Need to note, this is true for any atom,"},{"Start":"05:45.500 ","End":"05:49.770","Text":"not just for oxygen or carbon,"},{"Start":"05:49.770 ","End":"05:56.420","Text":"and that\u0027s why I haven\u0027t written specifically which atoms we\u0027re talking about."},{"Start":"05:56.420 ","End":"06:02.180","Text":"There\u0027s a useful equation that comes from the expression we\u0027ve just used."},{"Start":"06:02.180 ","End":"06:04.670","Text":"From inspection of the last equation,"},{"Start":"06:04.670 ","End":"06:06.620","Text":"we can see that N,"},{"Start":"06:06.620 ","End":"06:08.720","Text":"N is going to be the number of atoms,"},{"Start":"06:08.720 ","End":"06:10.980","Text":"is given by N,"},{"Start":"06:10.980 ","End":"06:17.405","Text":"the number of atoms is equal to the number of moles times Avogadro\u0027s number."},{"Start":"06:17.405 ","End":"06:20.000","Text":"This equation is very useful."},{"Start":"06:20.000 ","End":"06:24.725","Text":"We\u0027re going to use it extensively in the exercises."},{"Start":"06:24.725 ","End":"06:28.085","Text":"Now, we can also write this a bit more differently"},{"Start":"06:28.085 ","End":"06:32.135","Text":"by dividing each side by Avogadro\u0027s number."},{"Start":"06:32.135 ","End":"06:37.220","Text":"Then we get N over Avogadro\u0027s number,"},{"Start":"06:37.220 ","End":"06:41.675","Text":"number of atoms divided by Avogadro\u0027s number is equal to N,"},{"Start":"06:41.675 ","End":"06:43.865","Text":"the number of moles."},{"Start":"06:43.865 ","End":"06:45.950","Text":"This is the useful equation."},{"Start":"06:45.950 ","End":"06:50.585","Text":"We need to note that this is also true for molecules,"},{"Start":"06:50.585 ","End":"06:54.860","Text":"ions, anything, because we can really have a mole of anything."},{"Start":"06:54.860 ","End":"06:58.655","Text":"We can even have a mole of apples or oranges."},{"Start":"06:58.655 ","End":"07:04.490","Text":"In this video, we\u0027ve discussed the conversion factors that relate mole,"},{"Start":"07:04.490 ","End":"07:06.879","Text":"molar mass, and Avogadro\u0027s number."},{"Start":"07:06.879 ","End":"07:10.100","Text":"We\u0027ll find out that we can use these conversion factors to"},{"Start":"07:10.100 ","End":"07:14.820","Text":"solve a whole range of problems."}],"ID":17640},{"Watched":false,"Name":"Exercise 1","Duration":"3m 46s","ChapterTopicVideoID":22840,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22840.jpeg","UploadDate":"2020-12-10T05:08:31.1430000","DurationForVideoObject":"PT3M46S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.525","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.525 ","End":"00:06.270","Text":"What is the total number of atoms in: A,"},{"Start":"00:06.270 ","End":"00:08.880","Text":"17 mole of iron; B,"},{"Start":"00:08.880 ","End":"00:12.795","Text":"0.0065 mole of silver;"},{"Start":"00:12.795 ","End":"00:16.448","Text":"or C, 6 times 10 to the negative 12 mole of sodium?"},{"Start":"00:16.448 ","End":"00:18.105","Text":"We\u0027re going to start with A."},{"Start":"00:18.105 ","End":"00:20.430","Text":"We want to calculate the number of atoms."},{"Start":"00:20.430 ","End":"00:23.550","Text":"For this purpose, we\u0027re going to use the equation n is the number of"},{"Start":"00:23.550 ","End":"00:28.065","Text":"moles equal N divided by N_A."},{"Start":"00:28.065 ","End":"00:31.020","Text":"Now, N_A in the denominator is"},{"Start":"00:31.020 ","End":"00:35.580","Text":"Avogadro\u0027s number and N in the numerator is a number of entities."},{"Start":"00:35.580 ","End":"00:36.810","Text":"Now here in our question,"},{"Start":"00:36.810 ","End":"00:38.040","Text":"it\u0027s the number of atoms,"},{"Start":"00:38.040 ","End":"00:41.175","Text":"but in the future, it might be molecules and so on."},{"Start":"00:41.175 ","End":"00:46.960","Text":"Here, N is the number of atoms and N_A is Avogadro\u0027s number."},{"Start":"00:47.860 ","End":"00:51.665","Text":"Here it\u0027s going to be the number of atoms in 1 mole."},{"Start":"00:51.665 ","End":"00:54.740","Text":"We want to calculate the number of atoms."},{"Start":"00:54.740 ","End":"00:58.300","Text":"Therefore, where you can use the equation N,"},{"Start":"00:58.300 ","End":"01:02.310","Text":"the number of atoms of Fe of iron equals,"},{"Start":"01:02.310 ","End":"01:05.420","Text":"we\u0027re going to multiply both sides by Avogadro\u0027s number, n,"},{"Start":"01:05.420 ","End":"01:08.900","Text":"which is the number of moles times N_A, Avogadro\u0027s number."},{"Start":"01:08.900 ","End":"01:11.150","Text":"The number of moles we know is 17,"},{"Start":"01:11.150 ","End":"01:13.450","Text":"so this is 17 mole,"},{"Start":"01:13.450 ","End":"01:21.540","Text":"times Avogadro\u0027s number, which is 6.022, times 10^23."},{"Start":"01:21.540 ","End":"01:23.850","Text":"Since we\u0027re talking about atoms here,"},{"Start":"01:23.850 ","End":"01:25.960","Text":"it\u0027s atoms per mole,"},{"Start":"01:25.960 ","End":"01:29.165","Text":"because that\u0027s the number of atoms we have in 1 mole."},{"Start":"01:29.165 ","End":"01:30.830","Text":"The moles cancel out,"},{"Start":"01:30.830 ","End":"01:41.290","Text":"and this equals 1.02 times 10^25 atoms of iron of Fe."},{"Start":"01:41.290 ","End":"01:44.960","Text":"Our answer for A is the number of atoms of the iron equals"},{"Start":"01:44.960 ","End":"01:50.180","Text":"1.02 times 10^25 atoms of iron. Next, we\u0027re going to go on to B."},{"Start":"01:50.180 ","End":"01:54.740","Text":"In B, we have 0.0065 moles of silver."},{"Start":"01:54.740 ","End":"01:57.815","Text":"Now, again, we\u0027re going to use the same equation."},{"Start":"01:57.815 ","End":"01:59.750","Text":"Again, the number of atoms,"},{"Start":"01:59.750 ","End":"02:10.410","Text":"N of silver equals the number of moles of silver times Avogadro\u0027s number."},{"Start":"02:11.650 ","End":"02:22.429","Text":"Number of moles of silver equal 0.00065 moles of silver times,"},{"Start":"02:22.429 ","End":"02:29.015","Text":"again, Avogadro\u0027s number, which is 6.022 times 10^23."},{"Start":"02:29.015 ","End":"02:33.560","Text":"Here again, atoms per mole."},{"Start":"02:33.560 ","End":"02:38.835","Text":"The moles cancel out and this equals 3.91"},{"Start":"02:38.835 ","End":"02:46.340","Text":"times 10^20 atoms of silver."},{"Start":"02:46.340 ","End":"02:52.385","Text":"That\u0027s our answer for B, 3.91 times 10^20 atoms of silver."},{"Start":"02:52.385 ","End":"02:55.460","Text":"Now, we\u0027re going to go on to C. In C, again, N,"},{"Start":"02:55.460 ","End":"03:03.680","Text":"the number of atoms of sodium equal the number of moles of sodium times N_A,"},{"Start":"03:03.680 ","End":"03:05.135","Text":"which is Avogadro\u0027s number."},{"Start":"03:05.135 ","End":"03:11.142","Text":"This equals 6 times 10 to"},{"Start":"03:11.142 ","End":"03:20.055","Text":"the negative 12 moles of sodium times 6.022 times 10^23,"},{"Start":"03:20.055 ","End":"03:22.960","Text":"and again, atoms per mole."},{"Start":"03:23.750 ","End":"03:26.670","Text":"The moles cancel out,"},{"Start":"03:26.670 ","End":"03:36.879","Text":"and this equals 3.61 times 10^12 atoms of sodium."},{"Start":"03:38.060 ","End":"03:47.559","Text":"That is our answer for C. Thank you very much for watching."}],"ID":23649},{"Watched":false,"Name":"Exercise 2","Duration":"6m 50s","ChapterTopicVideoID":22838,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22838.jpeg","UploadDate":"2020-12-10T05:08:17.2670000","DurationForVideoObject":"PT6M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.765","Text":"Hi, in the following exercise we are going to determine A,"},{"Start":"00:03.765 ","End":"00:08.475","Text":"the number of moles of zinc in a 365 gram sample of zinc metal."},{"Start":"00:08.475 ","End":"00:13.275","Text":"B, the number of chromium atoms in 235 kilograms of chromium."},{"Start":"00:13.275 ","End":"00:17.250","Text":"C, the mass of 1,000,000 atom sample of metallic gold."},{"Start":"00:17.250 ","End":"00:20.280","Text":"So we\u0027re going to begin with A and in A,"},{"Start":"00:20.280 ","End":"00:24.930","Text":"we want to find the number of moles in the zinc and we know the mass of zinc sample."},{"Start":"00:24.930 ","End":"00:26.670","Text":"So we\u0027re going to use the equation n,"},{"Start":"00:26.670 ","End":"00:29.550","Text":"which is the number of moles, equals m,"},{"Start":"00:29.550 ","End":"00:32.715","Text":"which is mass divided by m_w,"},{"Start":"00:32.715 ","End":"00:34.320","Text":"which is the molar mass."},{"Start":"00:34.320 ","End":"00:42.911","Text":"N, number of moles of zinc equal the mass of zinc divided by the molar mass of zinc."},{"Start":"00:42.911 ","End":"00:45.115","Text":"The mass of zinc is given,"},{"Start":"00:45.115 ","End":"00:47.720","Text":"n equals 365 grams."},{"Start":"00:48.120 ","End":"00:51.085","Text":"We have to find the molar mass of zinc."},{"Start":"00:51.085 ","End":"00:55.540","Text":"The molar mass of zinc is found in the periodic table of elements and it equals"},{"Start":"00:55.540 ","End":"01:01.550","Text":"65.39 grams per mole."},{"Start":"01:01.680 ","End":"01:08.190","Text":"This equals 5.58 mole."},{"Start":"01:08.190 ","End":"01:10.510","Text":"Now, if we look at the units for a minute,"},{"Start":"01:10.510 ","End":"01:13.600","Text":"I want to remind you when you divide by a fraction,"},{"Start":"01:13.600 ","End":"01:16.090","Text":"meaning here we have grams divided by a fraction,"},{"Start":"01:16.090 ","End":"01:21.460","Text":"grams per mole, it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"01:21.460 ","End":"01:25.730","Text":"So this equals grams times mole for grams."},{"Start":"01:25.730 ","End":"01:28.490","Text":"The reciprocal of this fraction."},{"Start":"01:28.490 ","End":"01:33.215","Text":"The grams cancel out and we\u0027re left with mole."},{"Start":"01:33.215 ","End":"01:36.830","Text":"The moles of the zinc equal 5.58 mole."},{"Start":"01:36.830 ","End":"01:38.474","Text":"That is our answer for A."},{"Start":"01:38.474 ","End":"01:40.640","Text":"Now, we\u0027re going to continue with B."},{"Start":"01:40.640 ","End":"01:46.010","Text":"In B, we have to find the number of chromium atoms in 235 kilograms of chromium."},{"Start":"01:46.010 ","End":"01:48.710","Text":"So again, we know the mass of the chromium sample share."},{"Start":"01:48.710 ","End":"01:54.170","Text":"The first step is to use the equation n number of moles equals m,"},{"Start":"01:54.170 ","End":"01:57.305","Text":"which is the mass divided by the molar mass."},{"Start":"01:57.305 ","End":"02:02.850","Text":"This will give us the number of moles of the chromium."},{"Start":"02:02.850 ","End":"02:06.330","Text":"The next step will be to find the number of atoms of the chromium."},{"Start":"02:06.330 ","End":"02:08.240","Text":"First of all, we\u0027re going to find the moles."},{"Start":"02:08.240 ","End":"02:10.625","Text":"The moles of chromium,"},{"Start":"02:10.625 ","End":"02:15.305","Text":"equal the mass of chromium divided by the molar mass of chromium."},{"Start":"02:15.305 ","End":"02:21.070","Text":"The mass of chromium is given 235 kilograms."},{"Start":"02:21.610 ","End":"02:25.520","Text":"The molar mass is found in the periodic table of elements,"},{"Start":"02:25.520 ","End":"02:29.040","Text":"and that equals 52 grams per mole."},{"Start":"02:30.670 ","End":"02:34.700","Text":"Here in the units, since we have kilograms and grams,"},{"Start":"02:34.700 ","End":"02:38.030","Text":"it will be easier to have grams and grams."},{"Start":"02:38.030 ","End":"02:40.490","Text":"So we\u0027re going to multiply by a conversion factor,"},{"Start":"02:40.490 ","End":"02:44.550","Text":"which is 1000 grams in 1 kilogram."},{"Start":"02:44.550 ","End":"02:47.750","Text":"The kilograms are going to cancel out and we\u0027ll be left with grams."},{"Start":"02:47.750 ","End":"02:55.003","Text":"So this equals 4,519.23 mole."},{"Start":"02:55.003 ","End":"02:57.410","Text":"Against the mole, again,"},{"Start":"02:57.410 ","End":"03:00.760","Text":"the mole is the same as we calculated in 1;"},{"Start":"03:00.760 ","End":"03:02.615","Text":"dividing by a fraction,"},{"Start":"03:02.615 ","End":"03:05.150","Text":"the same as multiplying with the reciprocal of"},{"Start":"03:05.150 ","End":"03:08.915","Text":"the fraction and the units comes out to be mole."},{"Start":"03:08.915 ","End":"03:13.340","Text":"We know that the moles of chromium are 4,519.23 mole."},{"Start":"03:13.340 ","End":"03:16.340","Text":"The next step is to find that chromium atoms."},{"Start":"03:16.340 ","End":"03:22.895","Text":"In order to do this, we\u0027re going to use another equation and the number of moles equal n,"},{"Start":"03:22.895 ","End":"03:27.390","Text":"which is the number of atoms divided by N_A,"},{"Start":"03:27.680 ","End":"03:30.645","Text":"which is the Avogadro constant."},{"Start":"03:30.645 ","End":"03:32.440","Text":"Now, we\u0027re looking for the number of atoms."},{"Start":"03:32.440 ","End":"03:37.690","Text":"So we\u0027re going to multiply both sides by Avogadro\u0027s constant."},{"Start":"03:37.690 ","End":"03:40.840","Text":"N, the number of atoms equal n,"},{"Start":"03:40.840 ","End":"03:42.280","Text":"which is the number of moles,"},{"Start":"03:42.280 ","End":"03:45.910","Text":"times N_A, which is Avogadro\u0027s constant."},{"Start":"03:45.910 ","End":"03:48.220","Text":"The number of atoms of the chromium equal the moles of"},{"Start":"03:48.220 ","End":"03:50.860","Text":"the chromium times the Avogadro number."},{"Start":"03:50.860 ","End":"03:54.620","Text":"So this equals the moles that we calculated before,"},{"Start":"03:55.010 ","End":"03:59.140","Text":"400,519.23 mole times Avogadro\u0027s number,"},{"Start":"03:59.140 ","End":"04:05.545","Text":"which equals 6.022 times 10 to the 23 atoms per mole,"},{"Start":"04:05.545 ","End":"04:10.515","Text":"because in 1 mole we have this amount of atoms."},{"Start":"04:10.515 ","End":"04:20.800","Text":"The moles cancel out and we get 2.72 times 10 to the 27 atoms."},{"Start":"04:22.940 ","End":"04:26.448","Text":"That is our final answer for B."},{"Start":"04:26.448 ","End":"04:29.730","Text":"Now, we\u0027re going to continue to C. In C,"},{"Start":"04:29.730 ","End":"04:34.985","Text":"we have to find the mass of a 1000000 atoms sample of metallic gold."},{"Start":"04:34.985 ","End":"04:38.225","Text":"We know that we have 1,000,000 atoms in our sample."},{"Start":"04:38.225 ","End":"04:41.750","Text":"The first stage we\u0027re going to use is to find the number of moles."},{"Start":"04:41.750 ","End":"04:43.835","Text":"The number of moles equal like before,"},{"Start":"04:43.835 ","End":"04:47.870","Text":"n which is the number of atoms divided by N_A,"},{"Start":"04:47.870 ","End":"04:49.595","Text":"which is Avogadro\u0027s constant."},{"Start":"04:49.595 ","End":"04:51.260","Text":"In this case, the number of moles of"},{"Start":"04:51.260 ","End":"04:55.415","Text":"the gold equal the number of atoms of the gold which are given,"},{"Start":"04:55.415 ","End":"05:04.560","Text":"which is 1,000,000 atoms divided by Avogadro\u0027s constant,"},{"Start":"05:04.560 ","End":"05:12.270","Text":"which is 6.022 times 10 to the 23 atoms per mole."},{"Start":"05:12.270 ","End":"05:20.905","Text":"This equals 1.66 times 10 to the negative 18 mole."},{"Start":"05:20.905 ","End":"05:23.930","Text":"Again, we\u0027re dividing by a fraction."},{"Start":"05:23.930 ","End":"05:26.840","Text":"It\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"05:26.840 ","End":"05:31.490","Text":"So it\u0027s atoms divided by atoms per"},{"Start":"05:31.490 ","End":"05:41.110","Text":"mole equals atoms times mole per atom."},{"Start":"05:41.870 ","End":"05:45.480","Text":"Atoms cancel out and we\u0027re left with mole."},{"Start":"05:45.480 ","End":"05:48.935","Text":"We now know the number of moles of the gold sample."},{"Start":"05:48.935 ","End":"05:53.510","Text":"The next step is to find the mass of the gold sample of these 1,000,000 atoms."},{"Start":"05:53.510 ","End":"05:55.700","Text":"In order to find the mass, we\u0027re going to use the equation,"},{"Start":"05:55.700 ","End":"05:57.830","Text":"n number of moles equals m,"},{"Start":"05:57.830 ","End":"05:59.690","Text":"which is mass divided by m_w,"},{"Start":"05:59.690 ","End":"06:01.055","Text":"which is the molar mass."},{"Start":"06:01.055 ","End":"06:04.445","Text":"So we\u0027re going to multiply both sides by the molar mass to get the mass."},{"Start":"06:04.445 ","End":"06:07.550","Text":"The mass equals n,"},{"Start":"06:07.550 ","End":"06:10.570","Text":"which is the number of moles times m_w,"},{"Start":"06:10.570 ","End":"06:11.900","Text":"which is the molar mass."},{"Start":"06:11.900 ","End":"06:13.475","Text":"This is of course of gold,"},{"Start":"06:13.475 ","End":"06:16.040","Text":"and this equals the number of moles which we found here,"},{"Start":"06:16.040 ","End":"06:24.155","Text":"which we calculated 1.66 times 10 to the negative 18 moles times the molar mass of gold,"},{"Start":"06:24.155 ","End":"06:27.380","Text":"which is taken from the periodic table of elements,"},{"Start":"06:27.380 ","End":"06:35.160","Text":"equals 196.97 grams per mole."},{"Start":"06:35.160 ","End":"06:36.944","Text":"Moles, of course cancel out."},{"Start":"06:36.944 ","End":"06:45.310","Text":"This equals to 3.27 times 10 to the negative 16 grams."},{"Start":"06:45.310 ","End":"06:51.000","Text":"That is our final answer for C. Thank you very much for watching."}],"ID":23650},{"Watched":false,"Name":"Exercise 3","Duration":"4m 18s","ChapterTopicVideoID":22839,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22839.jpeg","UploadDate":"2020-12-10T05:08:24.7630000","DurationForVideoObject":"PT4M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.655","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.655 ","End":"00:08.775","Text":"How many silver atoms are present in a sample that contains 22.4 percent silver by mass."},{"Start":"00:08.775 ","End":"00:10.770","Text":"The sample weighs 80 grams."},{"Start":"00:10.770 ","End":"00:16.660","Text":"We know that the mass of our sample equals 80 grams."},{"Start":"00:16.940 ","End":"00:22.320","Text":"We also know that the sample contains 22.4 percent silver biomass."},{"Start":"00:22.320 ","End":"00:25.780","Text":"Meaning that if we have 100 gram sample,"},{"Start":"00:28.430 ","End":"00:33.610","Text":"22.4 grams of Silver."},{"Start":"00:33.620 ","End":"00:37.574","Text":"In order to find the mass of the silver,"},{"Start":"00:37.574 ","End":"00:46.890","Text":"the mass of the silver equals the mass of the sample times 22.4"},{"Start":"00:46.890 ","End":"00:57.760","Text":"grams of silver divided by 100 grams of sample."},{"Start":"00:58.390 ","End":"01:02.150","Text":"This equals again, the mass of the sample is 80 grams."},{"Start":"01:02.150 ","End":"01:12.410","Text":"This equals 80 grams of the sample times 22.4 grams of silver,"},{"Start":"01:12.410 ","End":"01:18.240","Text":"divided by 100 grams of the sample."},{"Start":"01:18.310 ","End":"01:21.350","Text":"The grams of sample cancel out,"},{"Start":"01:21.350 ","End":"01:29.045","Text":"and this equals 17.92 grams of silver."},{"Start":"01:29.045 ","End":"01:31.955","Text":"We found the mass of the silver."},{"Start":"01:31.955 ","End":"01:35.580","Text":"Now, we want to find the amount of silver atoms."},{"Start":"01:35.580 ","End":"01:37.655","Text":"Now that we have the mass of the silver,"},{"Start":"01:37.655 ","End":"01:39.860","Text":"we\u0027re going to go ahead and calculate the number of"},{"Start":"01:39.860 ","End":"01:42.350","Text":"moles of the silver and then calculate the atoms."},{"Start":"01:42.350 ","End":"01:43.850","Text":"I want to remind you that n,"},{"Start":"01:43.850 ","End":"01:45.680","Text":"the number of moles equals m,"},{"Start":"01:45.680 ","End":"01:48.980","Text":"which is mass divided by the molar mass."},{"Start":"01:48.980 ","End":"01:51.725","Text":"The number of moles of silver,"},{"Start":"01:51.725 ","End":"01:54.080","Text":"equal the mass of silver,"},{"Start":"01:54.080 ","End":"01:56.885","Text":"divided by the molar mass of silver."},{"Start":"01:56.885 ","End":"01:59.540","Text":"Now the mass of the silver, which we calculated,"},{"Start":"01:59.540 ","End":"02:07.040","Text":"it equals 17.92 grams divided by the molar mass of the silver."},{"Start":"02:07.040 ","End":"02:11.660","Text":"The molar mass of the silver is found in the periodic table of elements and it equals"},{"Start":"02:11.660 ","End":"02:17.735","Text":"107.87 grams per mole."},{"Start":"02:17.735 ","End":"02:19.550","Text":"When we divide these 2,"},{"Start":"02:19.550 ","End":"02:23.765","Text":"we get 0.166 mole."},{"Start":"02:23.765 ","End":"02:26.900","Text":"Now I just want you look for a second at the units we have"},{"Start":"02:26.900 ","End":"02:32.165","Text":"grams divided by a fraction divided by grams per mole."},{"Start":"02:32.165 ","End":"02:35.600","Text":"Now, just want to remind you that when we\u0027re dividing by a fraction,"},{"Start":"02:35.600 ","End":"02:39.500","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"02:39.500 ","End":"02:41.120","Text":"Meaning instead of grams per mole,"},{"Start":"02:41.120 ","End":"02:43.835","Text":"we multiply it by mole per gram."},{"Start":"02:43.835 ","End":"02:47.490","Text":"Gram cancel out and the units we\u0027re left with is mole."},{"Start":"02:47.490 ","End":"02:51.245","Text":"The number of moles of the silver equal 0.166 mole."},{"Start":"02:51.245 ","End":"02:54.485","Text":"Now the last step is converting these moles into atoms."},{"Start":"02:54.485 ","End":"02:56.690","Text":"For this purpose, we\u0027re going to use the equation n,"},{"Start":"02:56.690 ","End":"03:00.230","Text":"the number of moles equals N divided by Na."},{"Start":"03:00.230 ","End":"03:04.775","Text":"Na is Avogadro\u0027s number and n is the number of entities,"},{"Start":"03:04.775 ","End":"03:07.595","Text":"which in our case is the number of atoms."},{"Start":"03:07.595 ","End":"03:12.515","Text":"But in the future, you might see that it can equal molecules and so on."},{"Start":"03:12.515 ","End":"03:19.110","Text":"N is atoms and Na is Avogadro\u0027s number."},{"Start":"03:23.470 ","End":"03:26.345","Text":"It\u0027s actually atoms per mole,"},{"Start":"03:26.345 ","End":"03:29.480","Text":"n, the number of atoms of silver,"},{"Start":"03:29.480 ","End":"03:31.295","Text":"which we want to calculate,"},{"Start":"03:31.295 ","End":"03:36.350","Text":"equals the number of moles times Avogadro\u0027s number."},{"Start":"03:36.350 ","End":"03:38.860","Text":"I\u0027m just going to multiply both sides by Avogadro\u0027s number."},{"Start":"03:38.860 ","End":"03:40.940","Text":"This equals the number of moles which we calculated,"},{"Start":"03:40.940 ","End":"03:47.660","Text":"which is 0.166 mole times Avogadro\u0027s number,"},{"Start":"03:47.660 ","End":"03:55.890","Text":"which is 6.022, times 10 to the 23 atoms per mole."},{"Start":"03:57.440 ","End":"03:59.780","Text":"The moles cancel out,"},{"Start":"03:59.780 ","End":"04:08.430","Text":"and this equals 1 times 10 to the 23 atoms of silver."},{"Start":"04:09.100 ","End":"04:15.035","Text":"The number of atoms of silver equals 1 times 10 to the 23 atoms."},{"Start":"04:15.035 ","End":"04:16.325","Text":"That\u0027s our final answer."},{"Start":"04:16.325 ","End":"04:19.380","Text":"Thank you very much for watching."}],"ID":23651},{"Watched":false,"Name":"Exercise 4 - Part a","Duration":"3m 1s","ChapterTopicVideoID":22843,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22843.jpeg","UploadDate":"2020-12-10T05:11:03.8600000","DurationForVideoObject":"PT3M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.715","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.715 ","End":"00:04.410","Text":"Determine: A,"},{"Start":"00:04.410 ","End":"00:09.540","Text":"the number of krypton atoms in a 7.5 milligrams sample of krypton; B,"},{"Start":"00:09.540 ","End":"00:13.598","Text":"the molar mass and the identity of an element if the mass of a"},{"Start":"00:13.598 ","End":"00:18.495","Text":"4.3 times 10^22 atom sample of the element is 3.35 grams."},{"Start":"00:18.495 ","End":"00:19.815","Text":"We\u0027re gong to start with A,"},{"Start":"00:19.815 ","End":"00:27.820","Text":"and in A we\u0027re given the mass of the krypton which equals 7.5 milligrams,"},{"Start":"00:27.950 ","End":"00:32.235","Text":"and we\u0027re asked to find the number of atoms of krypton."},{"Start":"00:32.235 ","End":"00:35.435","Text":"What we\u0027re going to do is we\u0027re going to find the number of moles of the krypton,"},{"Start":"00:35.435 ","End":"00:39.575","Text":"and then multiply the number of moles of the krypton by"},{"Start":"00:39.575 ","End":"00:44.380","Text":"Avogadro\u0027s number in order to find the number of atoms in the krypton."},{"Start":"00:44.380 ","End":"00:49.160","Text":"Recall that the number of moles equals the mass divided by the molar mass."},{"Start":"00:49.160 ","End":"00:52.550","Text":"In our case, the number of moles of krypton"},{"Start":"00:52.550 ","End":"00:59.165","Text":"equals the mass of krypton divided by the molar mass of krypton."},{"Start":"00:59.165 ","End":"01:03.330","Text":"The mass is 7.5 milligrams"},{"Start":"01:03.430 ","End":"01:08.390","Text":"divided by the molar mass which is found in the periodic table of elements,"},{"Start":"01:08.390 ","End":"01:13.530","Text":"which equals 83.8 grams per mole."},{"Start":"01:13.580 ","End":"01:17.900","Text":"Now, since we have milligrams in our numerator and grams in our denominator,"},{"Start":"01:17.900 ","End":"01:20.200","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"01:20.200 ","End":"01:25.420","Text":"In 1 gram, we have 1,000 milligrams."},{"Start":"01:25.420 ","End":"01:31.055","Text":"The milligrams cancel out and the number of moles of the krypton"},{"Start":"01:31.055 ","End":"01:37.265","Text":"equals 8.95 times 10 to the negative 5 mole."},{"Start":"01:37.265 ","End":"01:40.955","Text":"That\u0027s the number of moles of the krypton that we found."},{"Start":"01:40.955 ","End":"01:44.765","Text":"Now, to find the number of atoms of the krypton, we\u0027re going to use n,"},{"Start":"01:44.765 ","End":"01:49.315","Text":"the number of moles, equals n divided by N_A."},{"Start":"01:49.315 ","End":"01:53.150","Text":"N_A is Avogadro\u0027s number and n could have a number of values,"},{"Start":"01:53.150 ","End":"01:55.325","Text":"for example, atoms, molecules, and so on."},{"Start":"01:55.325 ","End":"01:57.350","Text":"In our case, its atoms."},{"Start":"01:57.350 ","End":"02:00.487","Text":"So n, the number of atoms of krypton,"},{"Start":"02:00.487 ","End":"02:04.400","Text":"equals the number of moles of krypton times N_A,"},{"Start":"02:04.400 ","End":"02:06.200","Text":"which is Avogadro\u0027s number."},{"Start":"02:06.200 ","End":"02:11.750","Text":"This equals n, the number of moles of krypton which we found,"},{"Start":"02:11.750 ","End":"02:16.610","Text":"which equals 8.95 times 10 to"},{"Start":"02:16.610 ","End":"02:22.364","Text":"the negative 5 mole times Avogadro\u0027s number,"},{"Start":"02:22.364 ","End":"02:31.950","Text":"which equals 6.022 times 10^23 atoms per mole."},{"Start":"02:33.350 ","End":"02:38.810","Text":"The moles cancel out and we find that the number of krypton atoms equals"},{"Start":"02:38.810 ","End":"02:49.325","Text":"5.39 times 10^19 atoms of krypton."},{"Start":"02:49.325 ","End":"02:56.260","Text":"That is our answer for A, 5.39 times 10^19 atoms of krypton."},{"Start":"02:57.260 ","End":"02:59.340","Text":"That is our answer for A."},{"Start":"02:59.340 ","End":"03:01.720","Text":"Thank you very much for watching."}],"ID":23652},{"Watched":false,"Name":"Exercise 4 - Part b","Duration":"3m 21s","ChapterTopicVideoID":22841,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22841.jpeg","UploadDate":"2020-12-10T05:10:47.5770000","DurationForVideoObject":"PT3M21S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.415","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.415 ","End":"00:07.440","Text":"Determine A, the number of krypton atoms in a 7.5 milligrams sample of krypton."},{"Start":"00:07.440 ","End":"00:11.610","Text":"B, the molar mass and the identity of an element if the mass of a 4.3"},{"Start":"00:11.610 ","End":"00:15.645","Text":"times 10^22 atom sample of the element is 3.35 grams."},{"Start":"00:15.645 ","End":"00:19.725","Text":"A was recorded in the previous video, now we\u0027re going on to B."},{"Start":"00:19.725 ","End":"00:23.715","Text":"We want to calculate the molar mass and the identity of an element."},{"Start":"00:23.715 ","End":"00:27.360","Text":"We know the number of atoms and we know the mass for these number of atoms."},{"Start":"00:27.360 ","End":"00:30.030","Text":"First of all, we\u0027re going to calculate the number of moles of"},{"Start":"00:30.030 ","End":"00:32.580","Text":"the element from the number of atoms."},{"Start":"00:32.580 ","End":"00:36.465","Text":"Then the next step, we\u0027ll find the mass of 1 mole of this element,"},{"Start":"00:36.465 ","End":"00:43.870","Text":"N at the number of atoms equals 4.3 times 10^22 atoms."},{"Start":"00:44.060 ","End":"00:51.185","Text":"We know that the mass equals 3.35 grams."},{"Start":"00:51.185 ","End":"00:54.095","Text":"First of all, we\u0027re going to calculate the number of moles that we have."},{"Start":"00:54.095 ","End":"00:57.700","Text":"N, the number of moles equals N divided by N_A."},{"Start":"00:57.700 ","End":"01:02.420","Text":"N_A is Avogadro\u0027s number and N in this case is the number of atoms."},{"Start":"01:02.420 ","End":"01:07.060","Text":"Again, the number of moles of this element equals N divided by N_A,"},{"Start":"01:07.060 ","End":"01:13.660","Text":"N is given and equals 4.3 times 10^22 atoms."},{"Start":"01:16.520 ","End":"01:18.980","Text":"N_A is Avogadro\u0027s number,"},{"Start":"01:18.980 ","End":"01:26.870","Text":"so it equals 6.022 times 10^23 atoms per mole."},{"Start":"01:28.770 ","End":"01:35.335","Text":"The number of moles that we have for our element equals 0.07 moles."},{"Start":"01:35.335 ","End":"01:46.210","Text":"Now we know that the mass of 0.07 mole of our element equals 3.35 grams."},{"Start":"01:46.210 ","End":"01:50.080","Text":"Now, we can calculate the mass of 1 mole of our elements."},{"Start":"01:50.080 ","End":"01:53.650","Text":"The mass of 1 mole"},{"Start":"01:53.650 ","End":"02:00.475","Text":"equals 1 mole times,"},{"Start":"02:00.475 ","End":"02:04.510","Text":"we know that 3.35 grams is"},{"Start":"02:04.510 ","End":"02:11.390","Text":"the mass of 0.07 moles of element."},{"Start":"02:11.390 ","End":"02:14.120","Text":"Just going to call our element E and we want to calculate"},{"Start":"02:14.120 ","End":"02:20.640","Text":"the mass of 1 mole of E. The moles of E cancel out."},{"Start":"02:21.100 ","End":"02:30.040","Text":"This equals 47.86 grams for 1 mole of atom."},{"Start":"02:30.040 ","End":"02:33.845","Text":"That\u0027s the molar mass that we found for our element."},{"Start":"02:33.845 ","End":"02:36.950","Text":"Now, if you look at the periodic table of elements,"},{"Start":"02:36.950 ","End":"02:39.590","Text":"in order to find the identity of our element,"},{"Start":"02:39.590 ","End":"02:42.080","Text":"we can see that the molar mass of"},{"Start":"02:42.080 ","End":"02:51.725","Text":"titanium equals 47.88 grams per mole,"},{"Start":"02:51.725 ","End":"03:02.040","Text":"and the molar mass we found equals 47.86 grams per mole."},{"Start":"03:02.650 ","End":"03:07.790","Text":"Therefore, the identity of our element is titanium."},{"Start":"03:07.790 ","End":"03:18.365","Text":"The molar mass that we found equals 47.88 grams per mole and the element is titanium."},{"Start":"03:18.365 ","End":"03:20.045","Text":"That is our final answer for B."},{"Start":"03:20.045 ","End":"03:22.380","Text":"Thank you very much for watching."}],"ID":23653},{"Watched":false,"Name":"Exercise 5","Duration":"4m 34s","ChapterTopicVideoID":22842,"CourseChapterTopicPlaylistID":86811,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22842.jpeg","UploadDate":"2020-12-10T05:10:55.1970000","DurationForVideoObject":"PT4M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:03.869","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.869 ","End":"00:10.200","Text":"How many K-41 atoms are present in a piece of potassium weighing 320 milligrams?"},{"Start":"00:10.200 ","End":"00:15.720","Text":"The percent natural abundance of K-41 is 6.7 percent."},{"Start":"00:15.720 ","End":"00:18.420","Text":"First of all, we\u0027re going to calculate the number of"},{"Start":"00:18.420 ","End":"00:21.570","Text":"atoms in the whole potassium sample."},{"Start":"00:21.570 ","End":"00:26.880","Text":"For this purpose, we\u0027re going to calculate the number of moles of the potassium sample,"},{"Start":"00:26.880 ","End":"00:31.710","Text":"and then we\u0027re going to calculate the number of atoms of the potassium sample."},{"Start":"00:31.710 ","End":"00:34.580","Text":"Quick reminder, n, the number of moles,"},{"Start":"00:34.580 ","End":"00:38.255","Text":"equals the mass divided by the molar mass."},{"Start":"00:38.255 ","End":"00:41.030","Text":"The number of moles of the potassium sample,"},{"Start":"00:41.030 ","End":"00:42.875","Text":"we\u0027re just going to call it k,"},{"Start":"00:42.875 ","End":"00:48.920","Text":"equals the mass of k divided by the molar mass of k. The mass is given,"},{"Start":"00:48.920 ","End":"00:56.060","Text":"the mass of the sample is 320 milligrams divided by the molar mass of potassium,"},{"Start":"00:56.060 ","End":"00:58.310","Text":"which is given in the periodic table of elements."},{"Start":"00:58.310 ","End":"01:02.310","Text":"It equals 39.1 grams per mole."},{"Start":"01:02.870 ","End":"01:04.980","Text":"Here we have milligrams and grams."},{"Start":"01:04.980 ","End":"01:08.780","Text":"We\u0027re just going to multiply by a conversion factor to make this easier."},{"Start":"01:08.780 ","End":"01:14.410","Text":"So we\u0027re going to multiply by 1 gram for every 1,000 milligrams."},{"Start":"01:14.410 ","End":"01:16.750","Text":"The milligrams cancel out,"},{"Start":"01:16.750 ","End":"01:26.935","Text":"and this equals 8.18 times 10 to the negative 3 mole of the potassium sample."},{"Start":"01:26.935 ","End":"01:30.648","Text":"Now, I want to calculate the number of atoms of the potassium samples."},{"Start":"01:30.648 ","End":"01:33.100","Text":"For this purpose, we\u0027re going to use the equation n,"},{"Start":"01:33.100 ","End":"01:36.845","Text":"the number of moles, equals n divided by N_A."},{"Start":"01:36.845 ","End":"01:41.260","Text":"N_A is Avogadro\u0027s number and n is the number of entities."},{"Start":"01:41.260 ","End":"01:43.345","Text":"In our case, it\u0027s the number of atoms."},{"Start":"01:43.345 ","End":"01:47.590","Text":"Avogadro\u0027s number is atoms per mole in our case."},{"Start":"01:47.590 ","End":"01:50.785","Text":"Because it\u0027s the number of entities in 1 mole."},{"Start":"01:50.785 ","End":"01:57.460","Text":"The number of atoms of the potassium sample equals the number of moles of the potassium."},{"Start":"01:57.460 ","End":"02:00.395","Text":"Multiply both sides by Avogadro\u0027s number."},{"Start":"02:00.395 ","End":"02:02.990","Text":"This equals the number of moles of potassium,"},{"Start":"02:02.990 ","End":"02:10.440","Text":"which is 8.18 times 10 to the negative 3 mole"},{"Start":"02:10.440 ","End":"02:20.285","Text":"of potassium times 6.022 times 10^23 atoms per mole."},{"Start":"02:20.285 ","End":"02:22.980","Text":"That\u0027s Avogadro\u0027s number."},{"Start":"02:22.980 ","End":"02:32.580","Text":"This equals 4.93 times 10^21 atoms of potassium."},{"Start":"02:32.580 ","End":"02:37.220","Text":"Now we want to calculate the number of the K-41 atoms which are present"},{"Start":"02:37.220 ","End":"02:41.510","Text":"in the sample that we just calculated the number of atoms inside the sample."},{"Start":"02:41.510 ","End":"02:45.155","Text":"We didn\u0027t want to know how many of these atoms are K-41 atoms."},{"Start":"02:45.155 ","End":"02:50.660","Text":"We know that the percent natural abundance of K-41 equals 6.7."},{"Start":"02:50.660 ","End":"02:55.490","Text":"Now, the natural abundance of the K-41 is 6.7 percent."},{"Start":"02:55.490 ","End":"03:00.630","Text":"This means that for every 1,000 potassium atoms,"},{"Start":"03:02.680 ","End":"03:09.530","Text":"67 of these atoms are K-14."},{"Start":"03:09.530 ","End":"03:19.160","Text":"Therefore, the number of atoms of K-41 equals the number of atoms of"},{"Start":"03:19.160 ","End":"03:23.870","Text":"the potassium times 67 atoms of"},{"Start":"03:23.870 ","End":"03:33.660","Text":"K-41 divided by 1,000 potassium atoms."},{"Start":"03:36.280 ","End":"03:42.230","Text":"This equals the number of atoms of potassium in our potassium sample."},{"Start":"03:42.230 ","End":"03:47.090","Text":"We calculated it equals 4.93 times"},{"Start":"03:47.090 ","End":"03:55.420","Text":"10^21 atoms of potassium."},{"Start":"03:55.940 ","End":"03:58.273","Text":"This is times, again,"},{"Start":"03:58.273 ","End":"04:05.280","Text":"67 atoms of K-41"},{"Start":"04:05.280 ","End":"04:12.745","Text":"for every 1,000 atoms of potassium."},{"Start":"04:12.745 ","End":"04:22.650","Text":"The atoms of potassium cancel out and this equals 3.3 times 10^20 atoms."},{"Start":"04:24.790 ","End":"04:31.350","Text":"The number of atoms of K-41 that we found is 3.3 times 10^20 atoms."},{"Start":"04:31.350 ","End":"04:33.080","Text":"That is our final answer."},{"Start":"04:33.080 ","End":"04:35.760","Text":"Thank you very much for watching."}],"ID":23654}],"Thumbnail":null,"ID":86811}]

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1.1

5

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