|Q| = Ccal*delta T + m(water)*C(water)*delta T
= 1.55*(26.81-24.47) + 101.10*4.184*(26.81-24.47)
= 3.627 + 989.8
= 993.4 J
This is heat released by dissociation
So,
Q = -993.4 J
Molar mass of CaBr2,
MM = 1*MM(Ca) + 2*MM(Br)
= 1*40.08 + 2*79.9
= 199.88 g/mol
mass(CaBr2)= 1.89 g
use:
number of mol of CaBr2,
n = mass of CaBr2/molar mass of CaBr2
=(1.89 g)/(1.999*10^2 g/mol)
= 9.456*10^-3 mol
Now use:
delta H dis = Q/number of mol
= -993.4 J / 9.456*10^-3 mol
= -1.05*10^5 J/mol
= -105 KJ/mol
Answer: -105 KJ/mol
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