MCQ Collection
china-gaokao-chemistry MCQs
Practice china-gaokao-chemistry questions with answers and explanations.
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Correct Answer: C. Increasing pressure shifts a gaseous equilibrium toward the side with fewer gas moles
Explanation:
A pressure increase favours the side occupying less gaseous volume; a catalyst changes rate, not equilibrium position.
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Correct Answer: A. Across a period, effective nuclear charge generally increases and atomic radius tends to decrease
Explanation:
Across a period, increasing nuclear charge with similar shielding pulls valence electrons closer.
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Correct Answer: C. A covalent bond forms through sharing electron density between atoms
Explanation:
Covalent bonding arises from shared electron density, commonly between nonmetal atoms.
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Correct Answer: D. Oxidation is an increase in oxidation number and reduction is a decrease
Explanation:
Oxidation corresponds to electron loss/increased oxidation number; reduction is the opposite.
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Correct Answer: A. A catalyst lowers activation energy for both forward and reverse reactions
Explanation:
It provides a lower-energy pathway in both directions, speeding equilibrium without changing K.
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Correct Answer: B. C1H4
Explanation:
The requested molecular formula contains exactly 1 carbon and 4 hydrogen atoms.
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Correct Answer: C. The limiting reagent is consumed first and determines the maximum product amount
Explanation:
The limiting reagent runs out first under the balanced stoichiometric ratio and caps theoretical yield.
Correct Answer: 0.25
Explanation:
M=n/V=1/4=0.250 mol/L.
Correct Answer: 3
Explanation:
pH=−log₁₀[H⁺]=−log₁₀(1e-03)=3.0.
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Correct Answer: C. A covalent bond forms through sharing electron density between atoms
Explanation:
Covalent bonding arises from shared electron density, commonly between nonmetal atoms.
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Correct Answer: A. Oxidation is an increase in oxidation number and reduction is a decrease
Explanation:
Oxidation corresponds to electron loss/increased oxidation number; reduction is the opposite.
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Correct Answer: D. A catalyst lowers activation energy for both forward and reverse reactions
Explanation:
It provides a lower-energy pathway in both directions, speeding equilibrium without changing K.
