Students dissolve one sugar cube in 100 mL of water at 20 °C, 40 °C, and 60 °C and time each trial once. They stir vigorously only in the 60 °C trial. Which revision would best allow the class to draw a valid conclusion about the effect of temperature on dissolving time?

Stir at the same rate in every trial and repeat each temperature several times

Use a larger sugar cube at higher temperatures so the dissolving times stay measurable

Add a fourth trial at 80 °C while keeping the other procedures unchanged

Replace the stopwatch with a digital timer that reads to 0.01 s

Measured magnetic moments Sample Moment (Bohr magnetons) X 5.9 Y 4.9 Z 0 Three aqueous samples contain Fe²⁺, Fe³⁺, and Zn²⁺ in some order. The spin-only magnetic moment is about √[n(n+2)] for n unpaired electrons (√35 ≈ 5.9 and √24 ≈ 4.9). Which assignment matches the measured moments in the table? Assume the iron ions are high spin.

X is Fe³⁺, Y is Fe²⁺, and Z is Zn²⁺

X is Fe³⁺, Y is Zn²⁺, and Z is Fe²⁺

X is Fe²⁺, Y is Fe³⁺, and Z is Zn²⁺

X is Zn²⁺, Y is Fe²⁺, and Z is Fe³⁺

The ions S²⁻, Cl⁻, K⁺, and Ca²⁺ all have the electron configuration of argon. Which list orders them from smallest to largest ionic radius?

Ca²⁺ < K⁺ < Cl⁻ < S²⁻

K⁺ < Ca²⁺ < Cl⁻ < S²⁻

Ca²⁺ < K⁺ < S²⁻ < Cl⁻

S²⁻ < Cl⁻ < K⁺ < Ca²⁺

Students want to investigate how temperature affects the rate at which a piece of chalk reacts with vinegar. Which statement is a testable, falsifiable hypothesis for the investigation?

If vinegar is warmed from 20 °C to 40 °C, chalk will stop fizzing in less time

Chalk dissolves in vinegar because vinegar is an acid that reacts with it

Chalk will dissolve in vinegar at some temperature or another

Warm vinegar is generally better than cold vinegar at dissolving chalk pieces

Period 3 data Element Atomic radius (pm) IE₁ (kJ/mol) Na 186 496 Mg 160 738 Al 143 578 P 110 1012 Which explanation best accounts for the lower first ionization energy of Al than of Mg, even though Al has more protons and a smaller radius?

Al's outer electron is in a 3p subshell, higher in energy and shielded by the 3s electrons

Al has a larger radius than Mg, so its valence electron is farther from the nucleus

Mg has a half-filled 3p subshell that makes its electrons especially hard to remove

Al has fewer protons than Mg, so the attraction for its valence electron is weaker

Pauling electronegativities Element Value Si 1.9 P 2.1 C 2.5 S 2.5 Cl 3.0 Using the electronegativity values given, which bond is the most polar, and which atom carries the partial negative charge?

Si–Cl, with chlorine negative

P–Cl, with phosphorus negative

Praxis Chemistry (5246)

Free Praxis Chemistry 5246 Practice Questions

10 sample questions for the Praxis Chemistry (5246), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Nature and Impact of Science and Engineering, Principles and Models of Matter and Energy, Chemical Reactions and Periodicity, Chemical Composition, Bonding, and Structure, Solutions and Acid-Base Chemistry.

Question 1 · Nature and Impact of Science and Engineering
Students dissolve one sugar cube in 100 mL of water at 20 °C, 40 °C, and 60 °C and time each trial once. They stir vigorously only in the 60 °C trial.
Which revision would best allow the class to draw a valid conclusion about the effect of temperature on dissolving time?
- A.Use a larger sugar cube at higher temperatures so the dissolving times stay measurable
- B.Add a fourth trial at 80 °C while keeping the other procedures unchanged
- C.Stir at the same rate in every trial and repeat each temperature several times
- D.Replace the stopwatch with a digital timer that reads to 0.01 s
Show answer and explanation
Answer: C. Stir at the same rate in every trial and repeat each temperature several times
Stirring only in the 60 °C trial is a second variable that changes along with temperature, so it confounds the result. Holding stirring constant and repeating trials isolates temperature. A larger cube at higher temperatures adds yet another variable, and more precise timing or an extra temperature does not remove the confound.
Question 2 · Principles and Models of Matter and Energy
Measured magnetic moments
Sample Moment (Bohr magnetons)
X 5.9
Y 4.9
Z 0
Three aqueous samples contain Fe²⁺, Fe³⁺, and Zn²⁺ in some order. The spin-only magnetic moment is about √[n(n+2)] for n unpaired electrons (√35 ≈ 5.9 and √24 ≈ 4.9). Which assignment matches the measured moments in the table? Assume the iron ions are high spin.
- A.X is Fe³⁺, Y is Fe²⁺, and Z is Zn²⁺
- B.X is Fe³⁺, Y is Zn²⁺, and Z is Fe²⁺
- C.X is Fe²⁺, Y is Fe³⁺, and Z is Zn²⁺
- D.X is Zn²⁺, Y is Fe²⁺, and Z is Fe³⁺
Show answer and explanation
Answer: A. X is Fe³⁺, Y is Fe²⁺, and Z is Zn²⁺
Ionization removes the 4s electrons first, so Fe³⁺ is 3d⁵ with 5 unpaired electrons (moment √35 ≈ 5.9), Fe²⁺ is 3d⁶ with 4 unpaired electrons in the high-spin case (√24 ≈ 4.9), and Zn²⁺ is 3d¹⁰ with none, so it is diamagnetic. Placing the 4s electrons last is the common error that would give the wrong d counts.
Question 3 · Chemical Reactions and Periodicity
A student heats 2.46 g of a hydrate, MgSO₄·xH₂O, until the mass is constant and obtains 1.20 g of anhydrous MgSO₄ (molar mass 120 g/mol; H₂O = 18 g/mol). What is the value of x?
- A.5
- B.7
- C.6
- D.8
Show answer and explanation
Answer: B. 7
The water lost is 2.46 − 1.20 = 1.26 g, which is 0.07 mol. The anhydrous salt is 1.20/120 = 0.010 mol, so the mole ratio of water to salt is 7 : 1. The other values come from arithmetic slips in either mole conversion.
Question 4 · Chemical Reactions and Periodicity
The ions S²⁻, Cl⁻, K⁺, and Ca²⁺ all have the electron configuration of argon. Which list orders them from smallest to largest ionic radius?
- A.K⁺ < Ca²⁺ < Cl⁻ < S²⁻
- B.Ca²⁺ < K⁺ < S²⁻ < Cl⁻
- C.S²⁻ < Cl⁻ < K⁺ < Ca²⁺
- D.Ca²⁺ < K⁺ < Cl⁻ < S²⁻
Show answer and explanation
Answer: D. Ca²⁺ < K⁺ < Cl⁻ < S²⁻
In an isoelectronic series the electron count is fixed, so the radius is set by nuclear charge: more protons pull the same 18 electrons in more tightly. Ca²⁺ (20 protons) is smallest and S²⁻ (16 protons) is largest. Ordering by the parent atom's size, or reversing the sequence, ignores nuclear charge.
Question 5 · Chemical Reactions and Periodicity
Pure PCl₅ at 0.50 M decomposes by PCl₅ ⇌ PCl₃ + Cl₂. At equilibrium [Cl₂] = 0.10 M. What is K?
- A.4.0 × 10¹
- B.2.0 × 10⁻²
- C.2.5 × 10⁻¹
- D.2.5 × 10⁻²
Show answer and explanation
Answer: D. 2.5 × 10⁻²
The ICE table gives [PCl₃] = 0.10 M and [PCl₅] = 0.50 − 0.10 = 0.40 M. K = (0.10)(0.10)/(0.40) = 2.5 × 10⁻². Using the initial PCl₅ gives 2.0 × 10⁻², leaving out one product gives 2.5 × 10⁻¹, and inverting gives 40.
Question 6 · Nature and Impact of Science and Engineering
Students want to investigate how temperature affects the rate at which a piece of chalk reacts with vinegar.
Which statement is a testable, falsifiable hypothesis for the investigation?
- A.Chalk dissolves in vinegar because vinegar is an acid that reacts with it
- B.If vinegar is warmed from 20 °C to 40 °C, chalk will stop fizzing in less time
- C.Chalk will dissolve in vinegar at some temperature or another
- D.Warm vinegar is generally better than cold vinegar at dissolving chalk pieces
Show answer and explanation
Answer: B. If vinegar is warmed from 20 °C to 40 °C, chalk will stop fizzing in less time
A good hypothesis states a specific, measurable prediction relating an independent and a dependent variable, so evidence could show it is wrong. The other choices are vague, are explanations rather than predictions, or could not be shown false.
Question 7 · Principles and Models of Matter and Energy
Period 3 data
Element Atomic radius (pm) IE₁ (kJ/mol)
Na 186 496
Mg 160 738
Al 143 578
P 110 1012
Which explanation best accounts for the lower first ionization energy of Al than of Mg, even though Al has more protons and a smaller radius?
- A.Al's outer electron is in a 3p subshell, higher in energy and shielded by the 3s electrons
- B.Al has a larger radius than Mg, so its valence electron is farther from the nucleus
- C.Mg has a half-filled 3p subshell that makes its electrons especially hard to remove
- D.Al has fewer protons than Mg, so the attraction for its valence electron is weaker
Show answer and explanation
Answer: A. Al's outer electron is in a 3p subshell, higher in energy and shielded by the 3s electrons
Al loses an electron from the 3p subshell, which lies slightly higher in energy and is shielded by the 3s electrons. The data show Al is smaller than Mg and has more protons, which rules out the other explanations; Mg has no 3p electrons.
Question 8 · Chemical Reactions and Periodicity
How many hydrogen atoms are in 0.9 g of water? (H = 1, O = 16; use Avogadro's number = 6 × 10²³.)
- A.3 × 10²²
- B.1.2 × 10²³
- C.6 × 10²²
- D.5.4 × 10²³
Show answer and explanation
Answer: C. 6 × 10²²
0.9 g ÷ 18 g/mol = 0.05 mol of H₂O, which is 3 × 10²² molecules. Each molecule has two hydrogen atoms, so there are 6 × 10²² H atoms. The 3 × 10²² value counts molecules, and 5.4 × 10²³ multiplies mass by Avogadro's number without converting to moles.
Question 9 · Chemical Reactions and Periodicity
Pauling electronegativities
Element Value
Si 1.9
P 2.1
C 2.5
S 2.5
Cl 3.0
Using the electronegativity values given, which bond is the most polar, and which atom carries the partial negative charge?
- A.S–Cl, with sulfur negative
- B.C–Cl, with carbon negative
- C.P–Cl, with phosphorus negative
- D.Si–Cl, with chlorine negative
Show answer and explanation
Answer: D. Si–Cl, with chlorine negative
Bond polarity increases with the difference in electronegativity. Si–Cl has a difference of 1.1, larger than P–Cl (0.9), C–Cl (0.5), or S–Cl (0.5), and the more electronegative chlorine takes the partial negative charge. Choosing the less electronegative atom as negative reverses the dipole.
Question 10 · Chemical Reactions and Periodicity
A(g) + B(g) ⇌ 2C(g) has K = 4.0. Starting with 1.0 M A and 1.0 M B and no C, what is [C] at equilibrium?
- A.1.0 M
- B.0.50 M
- C.0.67 M
- D.2.0 M
Show answer and explanation
Answer: A. 1.0 M
With x as the change, [A] = [B] = 1 − x and [C] = 2x, so (2x)²/(1 − x)² = 4. Taking the square root gives 2x/(1 − x) = 2, so x = 0.50 and [C] = 1.0 M. Reporting x gives 0.50, and omitting the coefficient of C gives 0.67.

Measured magnetic moments
Sample	Moment (Bohr magnetons)
X	5.9
Y	4.9
Z	0

Period 3 data
Element	Atomic radius (pm)	IE₁ (kJ/mol)
Na	186	496
Mg	160	738
Al	143	578
P	110	1012

Pauling electronegativities
Element	Value
Si	1.9
P	2.1
C	2.5
S	2.5
Cl	3.0

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