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dsp73
3 years ago
12

3. the [OH-] of a solution with a [H+] of 9.3 x 10^-4 M

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

[OH-] = 1.08x10^-11 M

Explanation:

From the question given,

Concentration of Hydrogen ion, [H+]

= 9.3x10^-4 M

Concentration of Hydroxide ion, [OH-] =..?

The concentration of the hydrogen ion, [H+] and hydroxide ion, [OH-] are related with the following formula:

[H+] x [OH-] = 1x10^-14

Using the above formula, we can easily calculate the concentration of the hydroxide ion, [OH-] as follow:

[H+] x [OH-] = 1x10^-14

[H+] = 9.3x10^-4 M

9.3x10^-4 x [OH-] = 1x10^-14

Divide both side by 9.3x10^-4

[OH-] = 1x10^-14 /9.3x10^-4

[OH-] = 1.08x10^-11 M

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What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
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Solution : Given,

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N_2+3H_2\rightleftharpoons 2NH_3

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Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

Now put all the given values in this expression, we get

Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625

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4 years ago
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2 years ago
32.33 mL of 1.031 M potassium hydroxide were required to reach the endpoint of a titration of 50.00 mL of nitric acid. What was
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