General Chemistry Practice

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1.1Classify Common Forms of Matter

Exam I | Problem 1.1 |

Classify each sample as an element, compound, homogeneous mixture, or heterogeneous mixture:

\text{air}, \quad \text{salt water}, \quad \text{granite}

Solution

Air is a homogeneous mixture because its composition is fairly uniform throughout.

Salt water is also a homogeneous mixture because the salt is dissolved evenly in the water.

Granite is a heterogeneous mixture because its different minerals remain visible as separate parts.

1.2Count Particles in an Atom

Exam I | Problem 1.2 |

A neutral atom has atomic number $12$ and mass number $24$.

How many protons, neutrons, and electrons does it have?

Solution

The atomic number gives the number of protons, so the atom has $12$ protons.

The mass number is

$$ A = Z + N $$

$$ 24 = 12 + N $$

which gives $N = 12$ neutrons.

Because the atom is neutral, it also has $12$ electrons.

1.3Use Group Trends in the Periodic Table

Exam I | Problem 1.3 |

A main-group element is in Group 2.

How many valence electrons does it usually have, and why are Group 18 elements especially stable?

Solution

Group 2 elements usually have $2$ valence electrons.

Group 18 elements are especially stable because their outer electron shells are full.

1.4Read Subshell Capacities

Exam I | Problem 1.4 |

How many electrons can a $p$ subshell hold, and how many can a $d$ subshell hold?

Solution

A $p$ subshell holds $6$ electrons.

A $d$ subshell holds $10$ electrons.

1.5Use Periodic Trends

Exam I | Problem 1.5 |

Periodic Trends

Moving from left to right across a period, what happens to atomic radius and electronegativity?

Solution

Across a period, atomic radius decreases.

Across a period, electronegativity generally increases.

1.6Find Formal Charge in a Lewis Structure

Exam I | Problem 1.6 |

In the best Lewis structure for carbon dioxide, what is the formal charge on each atom?

Solution

The best Lewis structure is

\mathrm{O=C=O}

Carbon has four valence electrons and four bonding pairs around it, so its formal charge is $0$.

Each oxygen has two lone pairs and one double bond, which also gives a formal charge of $0$.

So every atom in the preferred structure has formal charge $0$.

1.7Predict Shape and Polarity

Exam I | Problem 1.7 |

What are the molecular shape and polarity of water, $H_2O$?

Solution

Oxygen has four electron groups around it: two bonding pairs and two lone pairs.

That gives a tetrahedral electron geometry and a bent molecular shape.

Because the bond dipoles do not cancel in a bent molecule, $H_2O$ is polar.

1.8Name Common Compounds

Exam I | Problem 1.8 |

Give the names of the following substances:

\mathrm{NaCl}, \quad \mathrm{CO_2}, \quad \mathrm{HCl(aq)}, \quad \mathrm{CaCO_3}

Solution

$\mathrm{NaCl}$ is sodium chloride.

$\mathrm{CO_2}$ is carbon dioxide.

$\mathrm{HCl(aq)}$ is hydrochloric acid.

$\mathrm{CaCO_3}$ is calcium carbonate.

1.9Balance a Combustion Reaction

Exam I | Problem 1.9 |

Balance the equation and identify the reaction type:

\mathrm{CH_4 + O_2 \rightarrow CO_2 + H_2O}

Solution

Start by balancing carbon and hydrogen:

\mathrm{CH_4 + O_2 \rightarrow CO_2 + 2H_2O}

Now count oxygen atoms on the product side: there are $4$ oxygen atoms total, so the reactant side needs $2$ oxygen molecules.

The balanced equation is

\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}

This is a combustion reaction because a hydrocarbon reacts with oxygen to form carbon dioxide and water.

1.10Use the Mole Concept

Exam I | Problem 1.10 |

How many molecules are in $0.25$ mol of water?

Solution

Use Avogadro's number:

0.25 \text{ mol} \times 6.022 \times 10^{23} \frac{\text{molecules}}{\text{mol}}

= 1.5055 \times 10^{23} \text{ molecules}

Rounded to three significant figures, the answer is

1.51 \times 10^{23} \text{ molecules}

Difficulty

Integrated

2.1Convert Mass to Particles

Exam II | Problem 2.1 |

A sample contains $18.0$ g of water.

How many moles and how many molecules is that?

Solution

The molar mass of water is $18.0$ g/mol, so

n = \frac{18.0\text{ g}}{18.0\text{ g/mol}} = 1.00\text{ mol}

Now convert moles to molecules:

1.00\text{ mol} \times 6.022 \times 10^{23} \frac{\text{molecules}}{\text{mol}} = 6.02 \times 10^{23}\text{ molecules}

2.2Find the Limiting Reagent

Exam II | Problem 2.2 |

For the reaction

\mathrm{2H_2 + O_2 \rightarrow 2H_2O}

suppose you start with $5.0$ mol of $H_2$ and $2.0$ mol of $O_2$.

Which reactant is limiting, and how many moles of water can form?

Solution

The reaction needs $2$ mol of $H_2$ for every $1$ mol of $O_2$.

If $2.0$ mol of $O_2$ are available, the reaction requires

2.0 \times 2 = 4.0\text{ mol }H_2

Since $5.0$ mol of $H_2$ are available, hydrogen is in excess and oxygen is limiting.

From the balanced equation, $1$ mol of $O_2$ produces $2$ mol of $H_2O$, so

2.0\text{ mol }O_2 \rightarrow 4.0\text{ mol }H_2O

2.3Solve an Ideal Gas Problem

Exam II | Problem 2.3 |

What volume does $0.50$ mol of gas occupy at $1.00$ atm and $300$ K?

Use $R = 0.0821\ \mathrm{L\cdot atm/(mol\cdot K)}$.

Solution

Use the ideal gas law:

$$ PV = nRT $$

Solve for volume:

V = \frac{nRT}{P}

Substitute the values:

V = \frac{(0.50)(0.0821)(300)}{1.00}

V = 12.315\text{ L}

Rounded appropriately,

V \approx 12.3\text{ L}

2.4Use Molarity and Dilution

Exam II | Problem 2.4 |

How many milliliters of a $2.00$ M stock solution are needed to prepare $100.0$ mL of a $0.500$ M solution?

Solution

Use the dilution relationship:

$$ M_1V_1 = M_2V_2 $$

Solve for the stock volume:

V_1 = \frac{M_2V_2}{M_1}

V_1 = \frac{(0.500)(100.0)}{2.00} = 25.0\text{ mL}

2.5Compute Heat Flow

Exam II | Problem 2.5 |

How much heat is required to warm $50.0$ g of water by $10.0^\circ$C?

Use $c = 4.18\ \mathrm{J/(g\cdot^\circ C)}$.

Solution

Use

q = mc\Delta T

Substitute the values:

$$ q = (50.0)(4.18)(10.0) $$

q = 2090\text{ J}

Because the temperature increases, the heat is positive.

2.6Predict a Shift in Equilibrium

Exam II | Problem 2.6 |

For the reaction

\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}

suppose the current reaction quotient is $Q = 0.10$ and the equilibrium constant is $K = 0.50$.

Which way will the reaction shift?

Solution

Compare $Q$ and $K$.

Since

$$ Q < K $$

the system has too few products relative to equilibrium.

The reaction shifts to the right, toward ammonia, to increase $Q$ until equilibrium is reached.

2.7Calculate pH from a Buffer

Exam II | Problem 2.7 |

A buffer has $[A^-] = 10[HA]$ and $\mathrm{p}K_a = 7.00$.

What is the pH?

Solution

Use the Henderson-Hasselbalch equation:

\text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)

Substitute the ratio:

\text{pH} = 7.00 + \log(10)

\text{pH} = 7.00 + 1.00 = 8.00

2.8Use a Rate Law

Exam II | Problem 2.8 |

For a reaction with rate law

\text{rate} = k[A]^2[B]

by what factor does the rate change if $[A]$ is tripled and $[B]$ is cut in half?

Solution

The rate changes by the factor

3^2 \times \frac{1}{2} = 9 \times \frac{1}{2} = 4.5

So the rate increases by a factor of $4.5$.

Difficulty

Applied

3.1Choose the Tonicity

Final | Problem 3.1 |

A red blood cell is placed in a solution that has a lower solute concentration than the cell.

What happens to the cell?

Solution

The outside solution is hypotonic relative to the cell.

Water moves into the cell by osmosis.

The cell swells and may burst if the concentration difference is large enough.

3.2Read a Gas Mixture

Final | Problem 3.2 |

Air has a total pressure of $760$ mmHg, and oxygen makes up $21\%$ of the mixture by mole fraction.

What is the partial pressure of oxygen?

Solution

Use Dalton's law:

P_i = X_i P_{total}

P_{O_2} = 0.21 \times 760

P_{O_2} = 159.6\text{ mmHg}

Rounded appropriately,

P_{O_2} \approx 160\text{ mmHg}

3.3Find Percent Yield

Final | Problem 3.3 |

An experiment has a theoretical yield of $20.0$ g and an actual yield of $15.0$ g.

What is the percent yield?

Solution

Use

\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Substitute the values:

\% \text{yield} = \frac{15.0}{20.0} \times 100 = 75.0\%

3.4Decide Protonation from pH and pKa

Final | Problem 3.4 |

A weak acid drug has $\mathrm{p}K_a = 8.0$ and is placed in blood at pH $7.4$.

Which form is more abundant, the protonated form or the deprotonated form?

Solution

Because

\text{pH} < \text{p}K_a

the protonated form predominates.

That is the key pH-to-pKa rule for weak acids.

3.5Identify the Anode and Cathode

Final | Problem 3.5 |

In a galvanic cell, one metal electrode loses mass because its atoms become ions.

Is that electrode the anode or the cathode, and where do the electrons go?

Solution

Loss of mass means oxidation is happening there.

Oxidation occurs at the anode, so that electrode is the anode.

The electrons flow away from the anode and toward the cathode.

Difficulty

Challenge

4.1Match Shape, Polarity, and Solubility

Final | Problem 4.1 |

Rank the following from most likely to least likely to dissolve in water:

\mathrm{NaCl}, \quad \mathrm{NH_3}, \quad \mathrm{CH_4}

Solution

Water is polar, so ionic and polar substances dissolve better than nonpolar substances.

$\mathrm{NaCl}$ is ionic, so it is the most water-soluble.

$\mathrm{NH_3}$ is polar and can hydrogen bond, so it is next.

$\mathrm{CH_4}$ is nonpolar, so it is the least soluble.

So the ranking is

\mathrm{NaCl} > \mathrm{NH_3} > \mathrm{CH_4}

4.2Use Q and K After a Disturbance

Final | Problem 4.2 |

For a reaction at equilibrium, a small amount of product is removed.

Immediately after the removal, is $Q$ greater than, less than, or equal to $K$, and which way does the reaction shift?

Solution

Removing product lowers the product concentration, so $Q$ becomes less than $K$.

To restore equilibrium, the reaction shifts to the right and makes more product.

4.3Update a Buffer After Adding Acid

Final | Problem 4.3 |

A buffer contains $0.10$ mol of $HA$ and $0.10$ mol of $A^-$ in $1.0$ L of solution, and $\mathrm{p}K_a = 7.00$.

Then $0.01$ mol of strong acid is added.

What is the new pH?

Solution

The added acid reacts with the base form:

A^- + H^+ \rightarrow HA

So the buffer amounts become:

A^-: 0.10 - 0.01 = 0.09\text{ mol}

HA: 0.10 + 0.01 = 0.11\text{ mol}

Use Henderson-Hasselbalch:

\text{pH} = 7.00 + \log\left(\frac{0.09}{0.11}\right)

\text{pH} \approx 7.00 + \log(0.818)

\text{pH} \approx 7.00 - 0.087 = 6.91

4.4Combine Stoichiometry and the Ideal Gas Law

Final | Problem 4.4 |

For the reaction

\mathrm{2H_2 + O_2 \rightarrow 2H_2O}

how many liters of $O_2$ are needed at $1.00$ atm and $298$ K to make $2.00$ mol of water?

Use $R = 0.0821\ \mathrm{L\cdot atm/(mol\cdot K)}$.

Solution

From the balanced equation, $1$ mol of $O_2$ produces $2$ mol of water.

So to make $2.00$ mol of $H_2O$, you need

1.00\text{ mol }O_2

Now use the ideal gas law:

V = \frac{nRT}{P}

V = \frac{(1.00)(0.0821)(298)}{1.00}

V = 24.5\text{ L}

General Chemistry Practice

Selected problems

Core formulas

Direct

1.1Classify Common Forms of Matter

1.2Count Particles in an Atom

1.3Use Group Trends in the Periodic Table

1.4Read Subshell Capacities

1.5Use Periodic Trends

1.6Find Formal Charge in a Lewis Structure

1.7Predict Shape and Polarity

1.8Name Common Compounds

1.9Balance a Combustion Reaction

1.10Use the Mole Concept

Integrated

2.1Convert Mass to Particles

2.2Find the Limiting Reagent

2.3Solve an Ideal Gas Problem

2.4Use Molarity and Dilution

2.5Compute Heat Flow

2.6Predict a Shift in Equilibrium

2.7Calculate pH from a Buffer

2.8Use a Rate Law

Applied

3.1Choose the Tonicity

3.2Read a Gas Mixture

3.3Find Percent Yield

3.4Decide Protonation from pH and pKa

3.5Identify the Anode and Cathode

Challenge

4.1Match Shape, Polarity, and Solubility

4.2Use Q and K After a Disturbance

4.3Update a Buffer After Adding Acid

4.4Combine Stoichiometry and the Ideal Gas Law