How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g・°C) as it changes temperature from 25.0°C to 76.0°C? A) 5.34 kJ B) 1190 kJ C) 1.19 kJ D) 5340 kJ

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1. How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g・°C) as it changes temperature from 25.0°C to 76.0°C?

A) 5.34 kJ
B) 1190 kJ
C) 1.19 kJ
D) 5340 kJ
 
2. How much heat (in kJ) will be absorbed by a 50.3 g piece of aluminum (specific heat = 0.930 J/g・°C) as it changes temperature from 23.0°C to 67.0°C?
 
3. What is the final temperature (in °C) of 150.1 g of water (specific heat = 4.18 J/g・°C) at 24.2°C that absorbed 950. J of heat?
 
4. What is the specific heat of a 75.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J of heat?
 
5. A 130.3 g piece of copper (specific heat 0.38 J/g・°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature (in °C) of the copper? (The specific heat of water is 4.18 J/g・°C).
 
6. Cl₂ is a stable diatomic molecule. It can be decomposed to form two Cl atoms as shown below. Cl₂ → 2Cl(g) Predict the change in enthalpy (∆H) of this reaction.
A) ∆H is (-); endothermic
B) ∆H is (+); endothermic
C) ∆H is (-); exothermic
D) ∆H is (+); exothermic
E) ∆H = 0; thermoneutral
 
 
7. The First Law of Thermodynamics tells us
A) that the energy of the universe is always increasing.
B) that the energy of the universe is a constant.
C) that the energy for a chemical reaction is a constant.
D) that a chemical reaction will always release energy.
 
 
8. For the chemical reaction Ca (s) + ½ O₂ (g) → CaO (s) ∆H° = -635 kJ/mol. What is the enthalpy (in kJ/mol) for the reaction 2 CaO (s) → 2 Ca (s) + O₂ (g)?
 
9. For the chemical reaction H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol. What is the enthalpy (in kJ/mol) for the reaction 3 H₂ (g) + 3 F₂ (g) → 6 HF (g)
 
10. Using the equations 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol Determine the enthalpy (in kJ/mol) for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)
 
 
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