By using the ICE table :
initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X
when Ka = (X) (X) / (0.2-X)
so by substitution:
4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6
and PH = -㏒[H+]
= -㏒ 10^-6
= 6
The heat released by the water when it cools down by a temperature difference AT
is Q = mC,AT
where
m=432 g is the mass of the water
C, = 4.18J/gºC
is the specific heat capacity of water
AT = 71°C -18°C = 530
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J
and this is the amount of heat released by the water.
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Explanation:
