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natta225 [31]
2 years ago
14

What is the change in enthalpy in joules when 5.44 x 10- mol of AgCl dissolves in water according to the following chemical equa

tion:
AgCl(s) rightarrow Ag+(aq)+ Cl-(aq) AH=65.5 kj
Chemistry
1 answer:
ra1l [238]2 years ago
4 0

Answer:

\Delta H=35.3J

Explanation:

Hello,

In this case, since the dissolution of silver chloride involves a change in enthalpy of 65.5kJ per 1 mole  when undergone, for 5.44x10⁻⁴ moles, the enthalpy change is:

\Delta H=65.5\frac{kJ}{1mol}*5.44x10^{-4}mol\\ \\\Delta H=0.0353kJ*\frac{1000J}{1kJ}\\ \\\Delta H=35.3J

Best regards.

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For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o
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Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

  • MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

          S                   S              2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

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⇒ S = 1.1696 E-3 M

  • NaF(s) → Na+(aq)  +  F-(aq)

        0.10M     0.10M        0.10M

  • MgF2(s) ↔ Mg2+(aq)  + 2F-(aq)

          S'                 S'              2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility  is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

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