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Oksana_A [137]
3 years ago
11

The bomb calorimeter in Exercise 102 is filled with 987g water. The initial temperature of the calorimeter contents is 23.32. A

1.056g sample of benzoic acid is combusted in the calorimeter. What is the final temperature of the calorimeter contents
Chemistry
1 answer:
Sholpan [36]3 years ago
4 0

Answer:

25.907°C

Explanation:

In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C

The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.

Thus:

-q_{combust} = q_{water} + q_{calori}

q_{combust} = heat of combustion of benzoic acid

q_{water} = heat energy released to water

q_{calori} = heat energy released to the calorimeter

Therefore,

-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})

1.056*26.42 = [0.987*4.18 + 6.66](T_{f} - 23.32)

27.8995 = [4.12566+6.660](T_{f} - 23.32)

(T_{f} - 23.32) = 27.8995/10.7857 = 2.587

T_{f} = 23.32 + 2.587 = 25.907°C

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Answer:

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Can someone help me understand what 1b is asking and how to answer it?
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Answer:

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0.4167\ \text{M}^{-1}\text{min}^{-1}

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We have the relation

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\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

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The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

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2 years ago
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