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Maurinko [17]
3 years ago
6

During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°

c. the specific heat of water is 4.18 j/g·°c. how much heat did the reaction generate?
Chemistry
1 answer:
pantera1 [17]3 years ago
6 0
Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
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The answer is 3.63. seconds.

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\frac{1}{[A]} =\frac{1}{[A]_{0} } +kt

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\[\begin{array}{l}{\rm{[A]  -  concentration\ of\ reactant\ A\ at\ time\ t}}\\{{\rm{[A]}}_0}{\rm{ -  initial\ concentration\ of\ reactant\ A}}\\{\rm{t - time}}\\{\rm{k  -  rate\ constant}}\end{array}\]

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k = 0.54 m^{-1}. s^{-1}

[NO_{2} ]= 0.62\ M

[NO_{2} ]_{0} = 0.28\ M

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\frac{1}{0.28\ M} =\frac{1}{0.62\ M } +(0.54 m^{-1}.s^{-1})  t\\\\(0.54 m^{-1}.s^{-1})  t= \frac{1}{0.28\ M} -\frac{1}{0.62\ M } = 1.959 \\\\

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t =\frac{1.959}{0.54} = 3.63\ seconds

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To learn more about second order reaction visit:

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