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gizmo_the_mogwai [7]
4 years ago
13

Explain why the boiling point of H2S with relative molecular mass 34 is lower than that of H2O with relative molecular mass 18​

Chemistry
1 answer:
Levart [38]4 years ago
4 0

Answer and Explanation:

In H2O molecules, due to formation of intermolecular hydrogen bonds, there is molecular association. Large amount of energy is required to break these intermolecular hydrogen bonds. Intermolecular hydrogen bonding is not possible in H2S. Hence, its boiling point is lower and is a gas.

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Go'=30.5 kJ/mol
makvit [3.9K]

Answer:

a) Keq = 4.5x10^-6

b) [oxaloacetate] = 9x10^-9 M

c) 23 oxaloacetate molecules

Explanation:

a) In the standard state we have to:

ΔGo = -R*T*ln(Keq) (eq.1)

ΔGo = 30.5 kJ/moles = 30500 J/moles

R = 8.314 J*K^-1*moles^-1

Clearing Keq:

Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6

b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])

4.5x10^-6 = ([oxaloacetate]/(0.20*10)

Clearing [oxaloacetate]:

[oxaloacetate] = 9x10^-9 M

c) the radius of the mitochondria is equal to:

r = 10^-5 dm

The volume of the mitochondria is:

V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L

1 L of mitochondria contains 9x10^-9 M of oxaloacetate

Thus, 4.18x10^-42 L of mitochondria contains:

molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules

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Explanation:

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