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Sadah_17 Sadah_17
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A year ago
Did I do it right? can someone show me how


A 6.55 g sample of aniline (C6H5NH2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25 kJ/°C. If the initial temperature was 32.9°C, use the information below to determine the value of the final temperature of the calorimeter.

4 C6H5NH2(l) + 35 O2(g) → 24 CO2(g) + 14 H2O(g) + 4 NO2(g) ΔH°rxn = -1.28 × 10^4 kJ
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#1
A year ago
Moles of aniline = mass/molar mass

= 6.55/93.13 = 0.0703318 mole

Heat released by the reaction--

4 mole aniline released 1.28*10^4 kJ heat so,

0​.0703318 mole aniline will release heat,

= (1.28*10^4)*0.0703318/4

= 225.06174 kJ

this heat is absorbed by calorimeter so,

225.06174 = 14.25 ( T - 32.9)

On solving, T = 48.69°C​​​​​

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