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3 years ago

5

Select all that are true

1 answer:

kipiarov [429]3 years ago

8 0

I think it’s B not quite sure ! Sorry

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3 years ago

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4 years ago

When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C.

AlladinOne [14]

Explanation:

1). The given data is as follows.

$T_{i} = 25.87^{o}C$ , $T_{f} = 38.13^{o}C$

C = 5.73 $kJ/^{o}C$

Hence, calculate the change in enthalpy of the reaction as follows.

$\Delta E_{rxn} = -C \times \Delta T$

= $-5.73 \times (38.13 - 25.87)^{o}C$

= -70.25 KJ

As, number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(6 \times 12 + 14 \times 1)}$

= 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

$\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}$

= $-3.90 \times 10^{3}$ kJ/mol

Thus, we can conclude that $\Delta E_{rxn}$ for the reaction in kJ/mol hexane is $-3.90 \times 10^{3}$ kJ/mol .

2). As we know that,

Number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(7 \times 12 + 8 \times 1)}$

= 0.017 mol

$\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}$

= $-3.91 \times 10^{3} \times 0.017$

= -65.875 kJ

As, $\Delta E_{rxn} = C \times \Delta T
$

-65.875 kJ = $-C \times (37.57 - 23.12)^{o}C$

C = 4.56 $kJ/^{o}C$

Hence, heat capacity of the bomb calorimeter is 4.56 $kJ/^{o}C$ .

3 0

4 years ago