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Juliette [100K]
3 years ago
12

Which factors help you identify the solute and the solvent in a solution? Select all that apply. In a solution, the solvent is p

resent in a greater amount. In a solution, the solute dissolves in a solvent. In a solution, the solute is the dissolving agent.
Chemistry
2 answers:
Dima020 [189]3 years ago
5 0
<h2>Answer:</h2>

The first two statements are correct about the solute and solvent which are:

  • <u><em>In a solution, the solvent is present in a greater amount. </em></u>
  • <u><em>In a solution, the solute dissolves in a solvent. </em></u>
<h3>Explanation:</h3>
  • Lets take an example of sugar water solution.
  • In this solution sugar is the solute because it is present in less amount in solution and it is the one which dissolves in solute.
  • Water is a solvent because a solvent is present in large quantity and it is the dissolving agent.
KatRina [158]3 years ago
4 0

The answers would be:

In a solution, the solvent is present in a greater amount.

In a solutions, the solute dissolves in a solvent.

In general, these are the best answers. The solute is what is being dissolved and the solvent is what dissolves. A solvent comes in greater amounts in a solution and it is the dissolving agent.

For example, sugar and water.

To make a sugar water solution, you will need to dissolve sugar in water. Sugar is the solute in this case because it is what is being dissolved. The water is the solvent, because it dissolves the sugar.

If you had more sugar than water, then you cannot make a solution.  

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Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

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Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

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This will give us a heterogeneous mixture where we can see 2 distinct phases , one containing the dissolved solid and the other having undissolved solid.

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