CHEM125 – CHAPTER 06 – BASIC HOMEWORK

CHAPTER 6 BASIC HOMEWORK

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Question 1

Match corresponding quantities. Help: Definition of energy EnergyChoose…Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck’s constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Kinetic energyChoose…Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck’s constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Photon energyChoose…Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck’s constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Conservation of energyChoose…Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck’s constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Potential energyChoose…Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck’s constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth

Question 2

5400 J of heat are added to a system. The surroundings do 2300 J of work on system. What is ΔE in J? Help: Internal energy

Explanation

ΔE = q + w = 5400 J + 2300 J =7700 J

Question 3

2200 J of heat are extracted from a system. The system does 1400 J of work on the surroundings. What is ΔE in J?

Explanation

ΔE = q + w = -2200 J -1400 J = -3600 J

Question 4

The temperature of the system is fixed. 617 J of heat are added to the system. What is the work in J done on the system? Help: Two equations for energy Additional hint given in feedback

Explanation

Because the temperature is constant, ΔE = 0. So, ΔE = q + w = 0. So w = -q.

Question 5

The work done on a system for a certain adiabatic process is 73 J. What is ΔE in J? Help: Adiabatic process Additional hint given in feedback

Explanation

Because the process is adiabatic, q=” 0.” So, ΔE = q + w = w.

Question 6

How much energy in kJ is required to heat a calorimeter whose heat capacity is 567.5 J from 23.5oC to 55.7oC? Help: Calorimeter and heat capacity

Explanation

Don’t forget to change to kJ. q= CΔT = (567.5 J) (55.7 – 23.5 oC) / 1000 =18.27

Question 7

What is the heat capacity of a calorimeter in J/oC if 352.1 J of heat raises its temperature by 7.9 oC?

Explanation

q=CΔT 352.1 = C (7.9o) 44.569 = C

Question 8

What is the heat capacity of 0.63 kg of water in J/oC? Help: Specific heat

Explanation

Don’t forget to change to g. C = m Cs = (630 g)(4.184) C= 2,635.92

Question 9

The specific heat of iron is 0.45 J/(goC). How much heat in J is added to a 9.54 g iron nail to raise its temperature from 22oC to 569.4oC? Help: Specific heat example

Explanation

q=mCsΔT q= (9.54 g)(0.45 )(569.4oC-22oC) q=2,349.9882

Question 10

A bowl of 292 g of water is placed in a microwave oven that puts out 874 watts (J/s). How long would it take in seconds to increase the temperature of the water from 12.0oC to 55.0oC? Use SF. Help: Heating time

Explanation

q=mCsΔT (292)(4.184)(43) 52,534.304 J/ (874 J/s) 60.107 s

Question 11

211.9 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 15.1oC. After 459.8 g of an unknown compound at 96.4oC is added, the equilibrium T is 26.9oC. What is the specific heat of the unknown compound in J/(goC)? Help: Specific heat calculation

Explanation

mCsΔT=mCsΔT (211.9 g)(4.184)(26.9-15.1)= (459.8g)x(96.4-26.9) 0.327379=x

Question 12

A person takes a complicated switchback trail to the top of a mountain. For which of the following cases would the person end up with the same potential energy or height? Help: State functions Select one or more:

Explanation

Aint about how fast I get there Aint about what’s waiting on the other side It’s the climb

Question 13

A process takes place at constant pressure. The volume changes and the temperature increases by 71.7oC. The heat capacity of the system at constant volume, CV, is 218.6 kJ/oC. What is ΔE in kJ? (Careful–pay attention to the units of the heat capacity.) Help: Energy is state function

Explanation

ΔE = CvΔT =(218.6)(71.7) =15,673.62

Question 14

For which of the following is the enthalpy of formation, ΔHfo = 0? Help: Enthalpy of elements Select one or more:

Explanation

Remember compounds are not elements.

Question 15

For which of the following reactions is the enthalpy change in the reaction, ΔHrxno, equal to ΔHfo for NH4OH(s)? Help: Definition of enthalpy of formation Select one:

Explanation

An equation for the enthalpy of formation must only create one mol of product and have all substances in their standard states.

Question 16

Calculate the enthalpy of reaction for the following reaction: 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) ΔHfo(Fe2O3(s)) = -824.2 kJ/mol ΔHfo(C(s)) = ? ΔHfo(Fe(s)) = ? ΔHfo(CO2(g)) = -393.5 kJ/mol

Help: Enthalpy of reaction example

Explanation

ΔH= (Sum)products – (Sum)reactants 3(-393.5 kJ/mol) – 2 (-824.2 kJ/mol) -1,180.5 + 1648.4 467.9 kJ/mol It is positive= Endothermic

Question 17

The previous reaction is Help: Definition of endothermic and exothermic reactions Select one:

Question 18

Calculate the enthalpy of reaction for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔHfo(Fe2O3(s)) = -824.2 kJ/mol ΔHfo(CO(g)) = -110.5 ΔHfo(Fe(s)) = ? ΔHfo(CO2(g)) = -393.5 kJ/mol Use SF.

Explanation

3(-393.5 kJ/mol) – (3(-110.5) + (-824.2)) -1,180.5 – (-331.5-824.2) -1180.5 + 1155.7 -24.8 kJ/mol It is negative = exothermic

Question 19

The previous reaction isSelect one:

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