Physics 1 Mechanics Waves and Thermodynamics

General Chemistry

To solve these problems, we will go through each step methodically, applying principles of thermodynamics and stoichiometry. ### Problem 1: Specific Heat Capacity of the Metal We will use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water, assuming no heat loss to the surroundings. Let's denote: - \( m_m \) as the mass of the metal - \( c_m \) as the specific heat capacity of the metal - \( \Delta T_m \) as the change in temperature of the metal - \( m_w \) as the mass of the water - \( c_w \) as the specific heat capacity of water (which is \( 4.184 \, \text{J/g}^\circ\text{C} \)) - \( \Delta T_w \) as the change in temperature of the water The heat lost by the metal is given by: \[ Q_m = m_m \cdot c_m \cdot \Delta T_m \] The heat gained by the water is given by: \[ Q_w = m_w \cdot c_w \cdot \Delta T_w \] Since \( Q_m = Q_w \), we can set the two equations equal to each other: \[ m_m \cdot c_m \cdot \Delta T_m = m_w \cdot c_w \cdot \Delta T_w \] Now we can plug in the values: - \( m_m = 5.00 \, \text{g} \) - \( m_w = 100.0 \, \text{g} \) - \( c_w = 4.184 \, \text{J/g}^\circ\text{C} \) - \( \Delta T_m = 100.0^\circ\text{C} - 28.5^\circ\text{C} \) - \( \Delta T_w = 28.5^\circ\text{C} - 25.0^\circ\text{C} \) \[ 5.00 \, \text{g} \cdot c_m \cdot (100.0^\circ\text{C} - 28.5^\circ\text{C}) = 100.0 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot (28.5^\circ\text{C} - 25.0^\circ\text{C}) \] \[ 5.00 \, \text{g} \cdot c_m \cdot 71.5^\circ\text{C} = 100.0 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot 3.5^\circ\text{C} \] \[ 5.00 \, \text{g} \cdot c_m \cdot 71.5 = 100.0 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot 3.5 \] \[ 357.5 \, \text{g}^\circ\text{C} \cdot c_m = 1464.4 \, \text{J} \] \[ c_m = \frac{1464.4 \, \text{J}}{357.5 \, \text{g}^\circ\text{C}} \] \[ c_m = 4.096 \, \text{J/g}^\circ\text{C} \] Therefore, the specific heat capacity of the metal is approximately \( 4.096 \, \text{J/g}^\circ\text{C} \). ### Problem 2: Balancing the Chemical Equation and Determining the Limiting Reactant First, we balance the chemical equation: \[ 2\text{Al} (s) + 3\text{Cl}_2 (g) \rightarrow 2\text{AlCl}_3 (s) \] The equation is already balanced with 2 moles of aluminum reacting with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. Next, we determine the limiting reactant: 1. Calculate the moles of aluminum (\( n_{\text{Al}} \)): \[ n_{\text{Al}} = \frac{m_{\text{Al}}}{M_{\text{Al}}} \] Where \( m_{\text{Al}} \) is the mass of aluminum and \( M_{\text{Al}} \) is the molar mass of aluminum (\( M_{\text{Al}} \approx 26.98 \, \text{g/mol} \)). \[ n_{\text{Al}} = \frac{10.0 \, \text{g}}{26.98 \, \text{g/mol}} \] \[ n_{\text{Al}} \approx 0.3706 \, \text{mol} \] 2. Calculate the moles of chlorine gas (\( n_{\text{Cl}_2} \)): \[ n_{\text{Cl}_2} = \frac{m_{\text{Cl}_2}}{M_{\text{Cl}_2}} \] Where \( m_{\text{Cl}_2} \) is the mass of chlorine gas and \( M_{\text{Cl}_2} \) is the molar mass of chlorine gas (\( M_{\text{Cl}_2} \approx 70.90 \, \text{g/mol} \)). \[ n_{\text{Cl}_2} = \frac{15.0 \, \text{g}}{70.90 \, \text{g/mol}} \] \[ n_{\text{Cl}_2} \approx 0.2116 \, \text{mol} \] 3. Determine the limiting reactant by comparing the mole ratio from the balanced equation with the mole ratio from the given masses: From the balanced equation, the mole ratio of \( \text{Al} \) to \( \text{Cl}_2 \) is \( 2:3 \). Therefore, for every 2 moles of \( \text{Al} \), 3 moles of \( \text{Cl}_2 \) are required. \[ \frac{n_{\text{Al}}}{2} = \frac{0.3706 \, \text{mol}}{2} = 0.1853 \, \text{mol} \] \[ \frac{n_{\text{Cl}_2}}{3} = \frac{0.2116 \, \text{mol}}{3} = 0.0705 \, \text{mol} \] Since \( 0.1853 \, \text{mol} \) of \( \text{Al} \) would require \( 0.1853 \times \frac{3}{2} = 0.27795 \, \text{mol} \) of \( \text{Cl}_2 \) and we only have \( 0.2116 \, \text{mol} \), chlorine gas is the limiting reactant. Therefore, chlorine gas (\( \text{Cl}_2 \)) is the limiting reactant in this reaction.

Chemistry

1. A 5.00 g sample of an unknown metal, heated to 100.0°C, is placed into 100.0 g of water at 25.0°C in a calorimeter. The final temperature of the water is 28.5°C. Calculate the specific heat capacity of the metal. (Assume no heat loss to the surroundings.) 2. Given the chemical reaction below, balance the equation and determine the limiting reactant when 10.0 g of aluminum reacts with 15.0 g of chlorine gas (Cl2): 2Al (s) + 3Cl2 (g) → 2AlCl3 (s)

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