Answer:
53 cm³
Explanation:
When the student used dilute sulfuric acid, the reaction was complete after 60 s, because no more hydrogen was formed in the next 10 s.
The reaction would go faster if the student used a more concentrated acid, but 53 cm³ of hydrogen is the most that would form.
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
Learn more: brainly.com/question/24245395
Taking the average of more measurements decreases random error of measurement
Taking the average of many measurements is the most effective way to reduce random errors in a measurement. Because the certainty of the results grows as the number of data does, Less risk of random errors means that the value is more certain. Fewer measurements lead to less reliable data collection, which raises the likelihood of random errors.
The complete question is
Which procedure(s) decrease(s) the random error of a measurement: (1) taking the average of more measurements: (2) calibrating the instrument; (3) taking fewer measurements? Explain
To learn more about random errors:
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0.86 moles of CO2 - i’m not sure if this is right but it should be something like that