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Inga [223]
3 years ago
13

575 394 9550 bFzN0S come study with me!

Chemistry
2 answers:
ch4aika [34]3 years ago
8 0

Answer:

Oo oki

Explanation:

Rina8888 [55]3 years ago
7 0

Answer:

im good thxs for offering tho!

Explanation:

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Hydrogen bonds or covalent but I’m sure it’s hydrogen
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What is the henry's law constant for co2 at 20∘c? express your answer to three significant figures and include the appropriate u
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Actually, Henry's Law is an empirical value. It means that it was not obtained out of raw calculations or correlations. This was gathered from experimental results. Hence, you can search its data. At standard temperature of 25°C (298 K),

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How many valence electrons would there be for the element with the<br> following orbital diagram?
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8 0
3 years ago
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

3 0
3 years ago
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