ACT Science Practice Question #1628 (Hard (610)) | Test Citadel
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ACT Science Difficulty: Hard (610)

Digital ACT Science Practice Question #1628

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Suppose a fifth trial were conducted under identical conditions using 0.05 g of MnO2 (half the mass used in Trial 3). The time required to collect 50 mL of O2 would most likely be: Passage Context (Chemical Kinetics): A chemistry research team investigated the decomposition rate of hydrogen peroxide (H2O2) into water and oxygen gas at 25°C under varying catalyst conditions. Four trials were conducted with 100 mL of 1.0 M H2O2: Table 1: - Trial 1: No catalyst | Initial rate: 0.02 mL O2/sec | Time to 50 mL O2: 2,500 sec - Trial 2: 0.10 g Fe2O3 | Initial rate: 1.15 mL O2/sec | Time to 50 mL O2: 44 sec - Trial 3: 0.10 g MnO2 | Initial rate: 8.40 mL O2/sec | Time to 50 mL O2: 6 sec - Trial 4: 0.10 g KI | Initial rate: 3.60 mL O2/sec | Time to 50 mL O2: 14 sec
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Socratic AI Engine Step-by-Step Derivation
0ms Precomputed
⚡ 15-Second Desmos Speed Hack

Decreasing catalyst mass increases reaction time. 6 sec < Time < 44 sec.

Masterclass Solution & Distractor Trap Analysis

With half the catalyst mass, the reaction rate will decrease compared to Trial 3 (0.10 g MnO2, 6 seconds), but will remain significantly catalyzed compared to Trial 2 (0.10 g Fe2O3, 44 seconds) or uncatalyzed (2,500 seconds). The time will ...

Distractor Analysis: Trap choice eliminates careless test-takers who confuse roots with coordinates...

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