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Alexxx [7]
3 years ago
14

A chemical engineer studying the properties of fuels placed 1.670 g of a hydrocarbon in the bomb of a calorimeter and filled it

with O2 gas. The bomb was immersed in 2.550 L of water and the reaction initiated. The water temperature rose from 20.00°C to 23.55°C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of reaction for combustion (qV) per gram of the fuel?
Chemistry
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

856,68 J/gram

Explanation:

We know that the variaton of temperature of 1K is equal to 1°C

so if the ΔT is 3.55°C, it is 3.55K.

so

403 J - 1 K

x - 3.55K

X= 403 * 3.55 = 1430,65 J

1.670 g - 1430,65 J

1 g - y

y = \frac{1430,65}{1.670} = 856,68 J

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The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

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Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

  • The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298}  = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

V_{2} = \frac{58752}{4023}\\\\

<h3>V₂ = 15 L </h3>

Therefore, the final volume of the gas is 15 L

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