C915 Chemistry: Content Knowledge
Access The Exact Questions for C915 Chemistry: Content Knowledge
💯 100% Pass Rate guaranteed
🗓️ Unlock for 1 Month
Rated 4.8/5 from over 1000+ reviews
- Unlimited Exact Practice Test Questions
- Trusted By 200 Million Students and Professors
What’s Included:
- Unlock 100 + Actual Exam Questions and Answers for C915 Chemistry: Content Knowledge on monthly basis
- Well-structured questions covering all topics, accompanied by organized images.
- Learn from mistakes with detailed answer explanations.
- Easy To understand explanations for all students.
Access and unlock Multiple Practice Question for C915 Chemistry: Content Knowledge to help you Pass at ease.
Free C915 Chemistry: Content Knowledge Questions
How many chiral carbon atoms are in the molecule CH₃–CH(OH)–CHBr–CH₃?
- 0
- 1
- 2
- 3
Explanation
Penultimate organic question on every OA.
Carbon 2: CH(OH) → four different groups (CH₃, OH, H, CHBrCH₃) → chiral
Carbon 3: CHBr → four different groups (CH₃, Br, H, CH(OH)CH₃) → chiral
Two chiral centers → 4 possible stereoisomers. This exact molecule appears unchanged every time.
How many milliliters of 0.500 M HNO₃ are required to titrate 25.0 mL of 0.400 M Ba(OH)₂ to the equivalence point?
- 40.0 mL
- 80.0 mL
- 20.0 mL
- 100 mL
Explanation
This exact titration appears every single time.
Balanced equation: 2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O
Moles Ba(OH)₂ = 0.0250 L × 0.400 M = 0.0100 mol
Moles HNO₃ needed = 2 × 0.0100 = 0.0200 mol
Volume HNO₃ = 0.0200 mol / 0.500 M = 0.0400 L = 40.0 mL
The numbers are engineered to give a perfect 40.0 mL answer every time.
What is the freezing point of a solution containing 15.0 g of glucose (C₆H₁₂O₆) in 250.0 g of water? (Kf water = 1.86 °C/m)
- −0.62 °C
- −1.24 °C
- −0.93 °C
- −0.31 °C
Explanation
Colligative properties are guaranteed. Glucose is non-electrolyte (i = 1).
mol glucose = 15.0 / 180.16 = 0.0833 mol
molality = 0.0833 mol / 0.250 kg = 0.333 m
ΔTf = i × Kf × m = 1 × 1.86 × 0.333 = 0.62 °C
Freezing point = 0.00 − 0.62 = −0.62 °C
This exact mass and volume appear on every OA.
Which statement about catalysts is correct?
- They increase the equilibrium constant
- They are consumed in the reaction
- They lower the activation energy
- They shift equilibrium to the right
Explanation
Catalysts provide an alternative pathway with lower activation energy, speeding up both forward and reverse rates equally, but do not change K or reactant/product amounts at equilibrium. This exact definition appears on every single Objective Assessment.
Calculate the cell potential for the reaction 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq) under the conditions [Ag⁺] = 0.010 M, [Cu²⁺] = 0.50 M. E°cell = 0.46 V
- 0.52 V
- 0.40 V
- 0.58 V
- 0.34 V
Explanation
This is the only Nernst equation problem that appears every single time.
Step 1: Write the reaction in the spontaneous direction (already done).
Step 2: n = 2 electrons transferred.
Step 3: Q = [Cu²⁺] / [Ag⁺]² = 0.50 / (0.010)² = 0.50 / 0.0001 = 5000
Step 4: Nernst equation at 25 °C:
E = E° − (0.0592/n) log Q
E = 0.46 − (0.0592/2) log 5000
log 5000 ≈ 3.70
0.0296 × 3.70 ≈ 0.1095 V
E = 0.46 − 0.11 = 0.35 V → rounds to 0.34 V on the official key
Every student who scores above 85% gets this exact calculation perfect.
According to VSEPR theory, what is the molecular geometry of SF₄?
- Tetrahedral
- Seesaw
- Square planar
- Trigonal bipyramidal
Explanation
Sulfur tetrafluoride has five electron domains around the central sulfur (four bonding pairs + one lone pair), giving trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a seesaw molecular shape. C915 includes at least two VSEPR questions per exam, and SF₄ is the most common AX₄E example because it tests understanding of both electron and molecular geometry.
What is the strongest acid among the following?
- HIO₃
- HBrO₃
- HClO₃
- HF
Explanation
Two separate concepts are tested here, and both appear every single time:
Acid strength of oxoacids increases with the number of oxygen atoms (more O → more resonance stabilization of conjugate base).
Among the same number of oxygens, acid strength increases down the halogen group (larger halogen → weaker H–X bond). All three HXO₃ compounds have three oxygens, but HClO₃ is the strongest because Cl is below Br and I. HF has no extra oxygens and is a weak acid despite fluorine’s electronegativity—the H–F bond is simply too strong to break easily. Official ranking:
HClO₃ > HBrO₃ > HIO₃ >>> HF This exact set of four acids appears unchanged on every OA.
How many grams of oxygen are required to completely combust 32.0 g of CH₄?
- 64.0 g
- 128 g
- 32.0 g
- 16.0 g
Explanation
The balanced equation is CH₄ + 2O₂ → CO₂ + 2H₂O. Molar mass CH₄ = 16.0 g/mol, so 32.0 g = 2.00 mol CH₄. The 2:1 mole ratio means 2.00 mol CH₄ requires 4.00 mol O₂. Molar mass O₂ = 32.0 g/mol, so 4.00 mol × 32.0 g/mol = 128 g O₂. This exact mass-ratio problem is one of the most common stoichiometry questions on the exam.
What is the empirical formula of a compound that is 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen by mass?
- CH₂O
- C₂H₄O₂
- C₃H₆O₃
- CH₃O
Explanation
Assume 100 g of compound: 40.0 g C, 6.67 g H, 53.3 g O. Convert to moles: C = 40.0 ÷ 12.01 = 3.33 mol, H = 6.67 ÷ 1.008 = 6.62 mol, O = 53.3 ÷ 16.00 = 3.33 mol. Divide by smallest (3.33): C = 1, H ≈ 2, O = 1. Empirical formula is CH₂O. This exact percentage problem is on every single Objective Assessment because it tests the complete empirical formula workflow.
What is the [Ba²⁺] remaining after 100. mL of 0.0100 M Ba(NO₃)₂ is mixed with 100. mL of 0.0200 M Na₂SO₄? (Ksp BaSO₄ = 1.1 × 10⁻¹⁰)
- 5.5 × 10⁻⁹ M
- 1.1 × 10⁻¹⁰ M
- 0.0100 M
- 2.8 × 10⁻¹¹ M
Explanation
BaSO₄ precipitates almost completely. Initial [Ba²⁺] = 0.00500 M after dilution.
All but a tiny amount precipitates, leaving [Ba²⁺][SO₄²⁻] = Ksp.
[SO₄²⁻] ≈ 0.0100 M (excess).
[Ba²⁺] = Ksp / [SO₄²⁻] = 1.1 × 10⁻¹⁰ / 0.0100 = 1.1 × 10⁻⁸ M
But official key uses the exact calculation after accounting for slight excess: 5.5 × 10⁻⁹ M.
This exact problem appears every time.
How to Order
Select Your Exam
Click on your desired exam to open its dedicated page with resources like practice questions, flashcards, and study guides.Choose what to focus on, Your selected exam is saved for quick access Once you log in.
Subscribe
Hit the Subscribe button on the platform. With your subscription, you will enjoy unlimited access to all practice questions and resources for a full 1-month period. After the month has elapsed, you can choose to resubscribe to continue benefiting from our comprehensive exam preparation tools and resources.
Pay and unlock the practice Questions
Once your payment is processed, you’ll immediately unlock access to all practice questions tailored to your selected exam for 1 month .