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Bumek [7]
3 years ago
11

PLEASE HELP ME!!!! ILL FIND A WAY TO GET YOU 100 POINTS!!!

Chemistry
1 answer:
Umnica [9.8K]3 years ago
8 0

It’s c had this problem last week

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A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion
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Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

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Heat of combustion = heat capacity of calorimeter * deltaT\\

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The heat of combustion = 10.5kJ/g.

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deltaT =heat  capacity of calorimeter   * (change in temperature)\\10.5kJ/g = heat  capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC

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