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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?

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VictorSilva

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 = [tex]\frac{1430,65}{1.670}[/tex] = 856,68 J

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