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mariarad [96]
3 years ago
7

A 1.11 g sample of caffeine, C8H10N4O2, burns in a constant-volume calorimeter that has a heat capacity of 7.85 kJ/K. The temper

ature increases from 298.70 K to 303.81 K. What is the molar heat of combustion of caffeine (in kJ).?

Chemistry
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

401135 kJ

Explanation:

From the balanced quation,

(q/n) = CΔE

Molar heat of combustion = 7.85kJk × (303.81-298.70)k

                                            = 7.85kj × 5.11

                                             = 40.1135kj

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Answer:

44.2 L

Explanation:

Use Charles Law:

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We have all the values except for V₂; this is what we're solving for. Input the values:

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From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

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It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

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