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dedylja [7]
3 years ago
5

Complete the following reaction :HC≡CH + HBr→? ​

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
3 0

hope it helps you .............

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When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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