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Homework Answers
1)heat absorbed = mass of substance x specific heat x rise in temperature
Given
heat absorbed =2913 J
mass = ??
specific heat = 5.19J/g.K
rise in temperature = final-initial temperautre = 315.8 -295.1=20.7 K
substituting the values
2913J = mass x 5.19J/g.K x 20.7K
Thus mass of substance = 27.11 g
2)
Kb of base = 6.0x10-6
Thus pKb of base = - log kb = -log 6.0x10-6 = 5.22
molarity of base C =0.33M
The pOH of a weak base is given by
pOH = 1/2[pKb-logC]
= 1/2[ 5.22 -log 0.33] =2.85
Thus pH of solutionof base = 14 - pOH = 14-2.85 =11.15
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to be given, please tell me what else you'd need because I don't
know.
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