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Gelneren [198K]
3 years ago
11

Consider the molar solubility of SrCO3 in 0.10 M Sr(NO3)2 versus in pure water. Ksp SrCO3 = 5.4 x 10-10 Which statement is true?

A. The solubility of Sr(NO3)2 increases. B. The molar solubility of SrCO3 will be higher in 0.10 M 0 M Sr(NO3)2. C. The molar solubility of SrCO3 will be lower in water. D. The molar solubility of SrCO3 will be lower in 0.10 M 0 M Sr(NO3)2 because of Sr2+ E. There is not enough information given to determine an answer
Chemistry
1 answer:
lesya692 [45]3 years ago
8 0

Answer:

The molar solubility of SrCO3 will be lower in 0.10 M 0 M Sr(NO3)2 because of Sr2+

Explanation:

The question reflects the phenomenon known in chemistry as common ion effect. Common ion effect refers to the decreased solubility of a solid in a solution that contains an ion in common with the solute.

In the case of our example here, both strontium carbonate and strontium nitrate possess the strontium II ion in common. The presence of this ion in strontium nitrate will decrease the solubility of strontium carbonate in accordance with Le Chateliers principle. Hence the answer.

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The atom has 7 neutrons and 8 protons....

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Sodium hydroxide + sulfuric acid​
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Answer:

NaOH + H2SO4 ‐‐> 2H2O + Na2SO4;

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Which statement best describes the variables in a controlled experiment?
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The statement that best describes the variables in a controlled experiment is independent variable while all other conditions are unchanged.

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8 0
3 years ago
Calculate the concentration of acetic acid, HAc, and acetate ion, Ac−, in a 0.25M acetate buffer solution with pH = 5.36. "0.25M
Neporo4naja [7]

Answer:

[HAc] = 0.05M

[Ac⁻] = 0.20M

Explanation:

The Henderson-Hasselbalch formula for the acetic acid buffer is:

pH = pka + log₁₀ [Ac⁻] / [HAc]

Replacing:

5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]

3.981 = [Ac⁻] / [HAc] <em>(1)</em>

Also, as total concentration of buffer is 0.25M it is possible to write:

0.25M =  [Ac⁻] + [HAc] <em>(2)</em>

Replacing (2) in (1)

3.981 = 0.25M - [HAc] / [HAc]

3.981 [HAc] = 0.25M - [HAc]

4.981 [HAc] = 0.25M

<em>[HAc] = 0.05M</em>

Replacing this value in (2):

0.25M =  [Ac⁻] + 0.05M

<em>[Ac⁻] = 0.20M</em>

I hope it helps!

7 0
4 years ago
Is the molecule H2+ stable?
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The antibonding orbital is empty. Thus, H2 is a stable molecule.
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