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densk [106]
3 years ago
6

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?

Chemistry
1 answer:
Alex Ar [27]3 years ago
6 0

____________________________________________________

Answer:

Your answer would be a). 2.0 × 10-9

____________________________________________________

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

3.31 x 10^-5 M

=

[BrO-]

or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

Then you would find ka:

(3.31 x 10^-5)^2/0.55 =2 x 10^-9

____________________________________________________

<em>-Julie</em>

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Answer:

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Explanation:

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This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.

NO₃⁻  + 1e⁻ →  NO₂

In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.

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8 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g)
lora16 [44]

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<u>Explanation:</u>

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<u>Initial:</u>                c             -

<u>At Eqllm:</u>         c-c\alpha      2c\alpha

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The expression of K_{c} for above equation follows:

K_{c}=\frac{[NO_2]^2}{[N_2O_4]}

Putting values in above equation, we get:

K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997

Hence, the value of equilibrium constant is 0.997

4 0
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