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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:

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Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴  1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

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