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andriy [413]
3 years ago
5

What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?

Chemistry
2 answers:
bekas [8.4K]3 years ago
6 0

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Over [174]3 years ago
5 0

Answer:

the answer is 8.02

Explanation:

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Explanation:

The solution of the lactic acd and sodium lactate is referred to as a buffer solution.

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.

Buffer solutions are generally known to resist change in pH values.

When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:

HA + OH- → A- + H2O.

Since the added OH- is consumed by this reaction, the pH will change only slightly.

The NaOH reacts with the weak acid present in the buffer sollution.

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Which of the following would allow you to find the volume of an irregular shaped solid object?
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If you had 100 grams of wax(C25H52), predict...
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The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp
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The artifact is 570 years old. That is, 5.7 × 10² years.

Explanation:

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Let the initial activity for fresh Carbon-14 be A₀

And the activity at any other time be A

The rate of radioactive decay is given by

dA/dt = - KA

dA/A = - kdt

Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)

In [(A₀/2)/A₀] = - k t(1/2)

In (1/2) = - k t(1/2)

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Integrating the left hand side from A₀ to A and the right hand side from 0 to t

In (A/A₀) = - kt

A/A₀ = e⁻ᵏᵗ

A = A₀ e⁻ᵏᵗ

A = 2.8 × 10³ Bq.

A₀ = 3.0 × 10³ Bq.

2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ

0.9333 = e⁻ᵏᵗ

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- kt = - 0.06899

t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years

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