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Tpy6a [65]
3 years ago
14

Which of these substances is the least acidic? solution A, pOH = 1.5 solution B, pOH = 7.0 solution C, pOH = 13.5Which of these

substances is the least acidic? solution A, pOH = 1.5 solution B, pOH = 7.0 solution C, pOH = 13.5
Chemistry
2 answers:
Yuliya22 [10]3 years ago
4 0

Answer:

A on ed

Explanation:

Usimov [2.4K]3 years ago
3 0

Answer: solution A, pOH =1.5

Explanation:

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How does Gibbs free energy predict spontaneity?
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2 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
Read 2 more answers
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