Multiple Choice

Topics: Atom, Electron configuration, Electron Pages: 6 (2103 words) Published: October 18, 2012
1. Which of the following statements about heat is false? A) If heat flows into a system, the extra energy of the system appears in the form of internal energy. B) A hot object possesses more heat than a cold object. C) If the system and surroundings are in thermal equilibrium, there is no heat flow between them. D) A process in which heat flows out of a system is said to be exothermic. E) Heat is a form of energy flow.

B is false. If you compare a cup of water at 25oC and a bath tub of water at 20oC, the cup of water may be warmer, but there are many fewer atoms than the bathtub of water, so there can actually be more heat in a colder object, but it is spread out throughout many more atoms so the temperature can be lower.

2. 2. Given: 60C(s) → C60; ΔH = 2320 kJ what is ΔH for the following thermochemical equation? 1/60C(s)>>>> C(s) A) +38.7 kJ B) +2320 kJ C) -129 MJ D) -2320 kJ E) -38.7 kJ

E is the correct answer. Since the forward reaction has a positive enthalpy, the reverse will have a negative enthalpy. Also we are dealing with 1/60th of the original amount, so (2320 kJ)*(1/60 mol)*(-1) = -38.7 kJ

3. Exactly 173.9 J will raise the temperature of 10.0 g of a metal from 25.0°C to 60.0°C. What is the specific heat capacity of the metal? A) 0.497 J/(g · °C) B) 2.01 J/(g · °C) C) 17.7 J/(g · °C) D) 41.6 J/(g · °C) E) none of these

A is the correct answer. Heat capacity is the energy it takes to raise the temperature of 1 gram of the substance by 1 oC. Since we raised 10 grams of the substance by 35 oC we can say that Heat capacity = 173.9 J/10 grams/35oC = 0.497 J/(g * oC)

4. In a bomb calorimeter, reactions are carried out A) at 1 atm pressure and 0°C. B) at a constant pressure. C) at a constant volume. D) at a constant pressure and 25°C. E) at 1 atm pressure and 25°C.

C is the correct answer. A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured. Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up the surrounding air, which expands and escapes through a tube that leads the air out of the calorimeter. When the air is escaping through the copper tube it will also heat up the water outside the tube. The temperature of the water allows for calculating calorie content of the fuel.

5. The standard enthalpy change for which of the following processes corresponds to the standard enthalpy of formation of solid rubidium chloride? A) Rb(s) + Cl2(s) → RbCl(s) B) Rb(g) + Cl2(g) → RbCl(g) C) Rb(g) + Cl(g) → RbCl(s) D) Rb(s) + Cl2(g) → RbCl(s) E) Rb(s) + Cl2(s) → RbCl(g)

D is the correct answer. The standard enthalpy of formation for elements in their standard state is zero. Rb(s) and Cl2(g) are in standard state so the reaction we are looking for is D

6. What is the standard enthalpy of formation of liquid butyraldehyde, CH3CH2CH2CHO(l)? CH3CH2CH2CHO(l) + O2(g) → 4H2O(l) + 4CO2(g); ΔH°= -2471.8 kJ Substance ΔH°ƒ(kJ/mol) CO2(g) -393.5 H2O(l) -285.8 A) –245.4 kJ/mol B) +245.4 kJ/mol C) –1792.5 kJ/mol D) –3151.1 kJ/mol E) +3151.1 kJ/mol

The answer is A. First we must balance the reaction for oxygen. There are 12 oxygens on the products side and 3 on the reactants side so we make (11/2) the coefficient for O2 on the reactants side to balance. Next we say that H(reaction) = Hf(products) – Hf(reactants) = 4*Hf(H2O) + 4*Hf(CO2) – (11/2) * Hf(O2) – Hf(CH3CH2CH2CHO)

so we rearrange the equation to give
Hf(CH3CH2CH2CHO) = 4mol *Hf(H2O) + 4mol *Hf(CO2) – (11/2) mol * Hf(O2) – H(reaction) = 4*(-285.8) + 4*(-393.5) – (11/2) * 0 – (-2471.8) = -245.4 kJ

7. Which of the following is not a fossil fuel? A) anthracite coal B) bituminous coal C) natural gas D) petroleum E) glucose

The answer is E....
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