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Sauron [17]
4 years ago
7

A teaspoon of dry coffee crystals dissolves when mixed in a cup of hot water. This process produces a coffee solution. Which ter

m best identifies how the original crystals are classified?
-Reactant
-product
-solvent
-a solute
Chemistry
2 answers:
Leviafan [203]4 years ago
7 0

Answer:

The correct answer is - solute.

Explanation:

A solute is one half of a solution that is dissolved in the other half of the solution, known as the solvent. Solute particles are always less or lower in quantity or concentration than the solvent in a solution.

In this case, coffee crystals are dissolved in hot water so the solute will be coffee crystals and the amount of coffee crystals are also less than the amount of solvent.

elena55 [62]4 years ago
6 0

Answer:

solute

Explanation:

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How many atoms are in 1.5 moles of calcium
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9.0×10^23

Explanation:

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Match each term to its example.
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3 years ago
A saturated solution of Barium Chloride at 30’C contains 150.0g water. How much additional Barium Chloride can be dissolved by h
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An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10-2M ) and calcium ion
Alona [7]

Answer : The remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, 1.07\times 10^{-6}M

Explanation :

The dissociation of barium fluoride is written as:

BaF_2\rightleftharpoons Ba^{2+}+2F^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ba^{2+}][F^-]^2

As we know that at room temperature 25^oC the K_{sp} of barium fluoride is, 1\times 10^{-6}.

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

1\times 10^{-6}=(2.75\times 10^{-2})\times [F^-]^2

[F^-]=6.03\times 10^{-3}M=0.00603M

The barium fluoride precipitate when fluoride ion is equal to 0.00603 M.

The dissociation of calcium fluoride is written as:

CaF_2\rightleftharpoons Ca^{2+}+2F^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][F^-]^2

As we know that at room temperature 25^oC the K_{sp} of calcium fluoride is, 3.90\times 10^{-11}.

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

3.90\times 10^{-11}=(6.70\times 10^{-2})\times [F^-]^2

[F^-]=2.41\times 10^{-5}M=0.0000241M

The calcium fluoride precipitate when fluoride ion is equal to 0.0000241 M.

Since, the fluoride ion concentration in calcium fluoride is less then the fluoride ion concentration in barium fluoride. That means, calcium fluoride will precipitate first.

Thus, the concentration Ca^{2+} ion remaining at F^- concentration (0.00603 M) is calculated as:

K_{sp}=[Ca^{2+}][F^-]^2

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

3.90\times 10^{-11}=[Ca^{2+}]\times (0.00603)^2

[Ca^{2+}]=1.07\times 10^{-6}M

Therefore, the remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, 1.07\times 10^{-6}M

3 0
3 years ago
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