College Chemistry Unit 8 Practice Test

The temperature of the cooling water as it leaves the hot engine of an automobile is 220 °F. After it passes through the radiator it has a temperature of 145 °F. Calculate the amount of heat transferred from the engine to the surroundings by 2 gallons of water with a specific heat of 4.184 J/g °C (1 gal = 3.785 L)

• 757 kJ

• 1627 kJ

• 1318 kJ

• 2375 kJ

Answer explanation

The key here is to start by finding the mass of the water. Remember 1 mL of water is 1g or 1 L of water is 1kg.

The mass must be in grams before solving for Q since the Cp of water is 4.184 J/g^oC.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 25.0 g of water at a temperature of 10.0°C. After stirring, the final temperature of both copper and water is 25.0°C. Assuming no heat losses, what is the mass of the piece of copper?

The specific heat capacity of water is 4.18 J/g·°C, the specific heat capacity of copper is 0.385 J/g·°C.

• 54.3 g

• 36.2 g

• 12.3 g

• 15.0 g

Answer explanation

Q = -Q problem.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2

NGSS.HS-PS3-4

When a 0.650-g sample of trinitrotoluene (TNT), C₇H₅N₂O₆, is burned in a bomb calorimeter, the temperature increases from 22.1 °C to 24.7 °C. The heat capacity of the calorimeter is 354 J/°C, and it contains 475 mL of water. To three significant figures, how much heat (kJ) was produced by the combustion of the TNT sample?

Mathematical Equivalence

ON

Answer explanation

Find out how much heat the water must have absorbed to warm up then find out how much the calorimeter must have absorbed to do the same. Add both of these numbers together.

Tags

NGSS.HS-PS3-2

NGSS.HS-PS3-4

The pictured sequence of reactions occurs in the commercial production of aqueous nitric acid:

Determine the total energy change for the production of one mole of aqueous nitric acid by this process.

Round your answer to 3 Sig Figs.

Mathematical Equivalence

ON

Answer explanation

This is extremely similar to a problem from the test. The trick is that each equation has to occur a certain number of times in order to properly balance all of them. Once done multiplying and adding them all together I will discover that I have found the energy released when 8 moles of HNO₃ are produced. I will simply divide that by 8.

Tags

NGSS.HS-PS1-4

Calculate ΔH°₂₉₈  for the process

Co₃O₄ (s) ⟶ 3Co(s) + 2O₂ (g)

from the following information:

Co(s) + O₂ (g) ⟶ CoO(s)     ΔH°₂₉₈  = −327.9kJ

3CoO(s) + O₂ (g) ⟶ Co₃O₄ (s)     ΔH°₂₉₈  = −189.5kJ

Mathematical Equivalence

ON

Answer explanation

Use the smaller equations pieced together like a puzzle to solve the bigger equation.

In this case both equations will be flipped changing the sign to positive. The first equation given will also need to occur 3 times meaning I need to multiply the ΔH₂₉₈ by 3.

Tags

NGSS.HS-PS3-1

When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies (in the instructions above) to predict the correct structure of the carbon dioxide molecule:

C-O is 350 kJ/mol

C=O is 741 kJ/mol

C=O is 1080 kJ/mol

• A is more stable with a total bond energy of 1482 kJ/mol

• B is more stable with a total bond energy of 1430 kJ/mol

• B is more stable with a total bond energy of 1482 kJ/mol

• A is more stable with a total bond energy of 1430 kJ/mol

Answer explanation

The molecule with a higher bond energy is a more stable configuration. This can also be proven with formal charges.

Tags

NGSS.HS-PS1-2

NGSS.HS-PS1-4

Rank these compounds in order from greatest to least energy required to convert one mole of the solid into separate ions.

• BeO

• MgF

• CaBr

• NaF

• RbBr

Answer explanation

First we compare the difference in their ionic charges. The greater the difference the strong the attraction. I like to visualize a number line. So for example +2 with -2 is stronger than a +2 with a -1 but still weaker is a +1 and -1.

To break ties where the charges are the same, larger atoms (further down the table) cause the bonds to weaken. The charges are farther apart weakening their attraction.

Tags

NGSS.HS-PS1-3

NGSS.HS-PS1-4

The melting point of H2S should be ​ (a)   than H2O due to the lack of ​hydrogen bonding.

• less
• greater

Answer explanation

Hydrogen bonding is very strong in relation to other IMF. This means boiling points are high. Like water.

H2S does not experience hydrogen bonding since we don't have NOF so we must assume that it will only experience dispersion forces and have a much lower boiling point.

On large covalent molecules like C₄H₉OH and CH₂(OH)CH₂OH what causes a change in boiling points other than molar mass?

• more polar locations raises boiling points

• increased hydrogen bonding raises boiling points

• increased hydrogen bonding lowers boiling points

• more polar locations lowers boiling points

Categorize the following compounds by the intermolecular forces that most strongly affect them.

Groups:

(a) Dispersion Forces

,

(b) Dipole-Dipole

,

(c) Hydrogen Bonding

• CH₃COOH

• HF

• NH₃

• NO

• SO₂

• CO

• CH₄

• C₃H₆

• Br₂

Answer explanation

Hydrogen bonding H-NOF.

Ion-Dipole. I need to be given and ionic or an ionic substance that will dissociate in water.

Dipole-Dipole. This is going to occur if it doesn't fit hydrogen bonding but is polar.

Dispersion. This is going to occur if it doesn't fit hydrogen bonding but is nonpolar.