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Mrrafil [7]
3 years ago
7

Propionic acid (CH3CH-COOH) has a K, 1.3 x 10^-5 and phosphoric acid (H3PO2) has a Ka = 7.5 x 10^-3

Chemistry
1 answer:
tatuchka [14]3 years ago
5 0

Answer:

D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4.

Explanation:

Hello!

In this case, since the ionization of a weak acid produces a conjugate acid, molecule to which the leaving H⁺ is added to, and a conjugate base, everything that remains after the withdrawal of the H⁺, we can note down the ionization of each acid down below:

CH_3CH_2COOH\rightleftharpoons CH_3CH_2COO^-+H^+\\\\H_3PO_4\rightleftharpoons H^++H_2PO_4^-

Thus, we notice that the answer is D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4 because the phosphoric acid is most likely to ionize to dihydrogen phosphate ions rather than phosphate, or hydrogen phosphate ions it undergoes a stepwise ionization in which the first step is the predominant one.

Best regards!

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Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

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An aqueous solution of glucose has molality of 6.81 and a density of 1.2 g/ml what is the molarity of glucose in the solution
andriy [413]

Answer:

the molarity is 3.68 moles/L

Explanation:

the molality of the solution of sucrose is

m= moles of glucose / Kg of solvent (water)= 6.81 ,

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then the molarity M is

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M= 3.68 moles/L

Note:

- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.

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