So I had this assignment, and I can\'t figure out the question number 2. I post

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So I had this assignment, and I can't figure out the question number 2. I post it here, and people keep telling me there is some more information needed to solve question 2, but this is the whole assignment I got and I was thinking maybe the information are in question 1?

I need the answer with the question 2, I only post question 1 because if the information needed is in that question!

Thanks!

1. To determine the heat capacity of a calorimeter you fill it with 470.92 ± 0.02 g water before igniting 0.107 t 0.001 g benzoic acid in it. The ignition is initiated with an ignition wire from which 8+1 g. burn in the process. The heat of combustion (AH2) of benzoic acid is-3228 ± 4 kJ/mol, and that of the ignition wire is-5.9 ± 0.4 kJ/g. The temperature of the system (calorimeter and water) rises by 1.357 ± 0.001 °C Calculate the heat capacity of the calorimeter. -Givc thc answcr with thc appropriatc accuracy (significant numbcrs) and with the correct error margins -S how all equations 2. To determine the heat of combustion for naphthalene a calorimeter is used that has a heat capacity (Col) of 250 ± 80 J/K. The calorimeter is ill with 47579 ± 0.02 g water before igniting 0.1090.001 g naphthalene in it. The ignition is initiated with an ignition wire from which 2+1 g. burn in the process. The heat of combustion (AH2) of the ignitior wire is-5.9 ± 0.4 kJ/g. The temperature of the system (calorimeter and water) rises by 1.853 ±0001°C Calculate the heat of combustion of napthalene -Give the answer with the appropriate accuracy (significant numbers) and. with the correct error margins Show all equations

Explanation / Answer

2. The heat of combustion (hereinafter referred to HC0) of a substance is defined as the thermal effect that accompanies a combustion process.

Qin = Qout

Q(comb) + Q ignition of wire = -(q(water) + q(bomb)]

Q combustion = mass of water X specific heat of water X change in temperature + heat capacity of calorimeter - Q ignition of wire

Qcomubstion = 475.79 X 4.18 X 1.853 + 250 X 1.853 + 5900X 2

Q comubstion = 3685.25 + 463.25 +11800 = 14985.25 Joules

so this is for 0.109 grams of naphthalene

0.109 grams = 0.109 / 128 moles = 0.000851 moles

so for 0.000851 moles heat of comustion is 14985.25 Joules

so for one mole = 14985.25 / 0.000851 = 17608.98 Kj / moles

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