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slamgirl [31]
3 years ago
13

A chemistry student adds a quantity of an unknown solid compound X to 5.00 L of distilled water at 15.° C . After 10 minutes of

stirring, only a little X has dissolved. The student then drains off the solution and evaporates the water under vacuum. A precipitate is left behind. The student washes, dries and weighs the precipitate. It weighs 0.17 kg.
Can you calculate the solubility of X in water at 15.° C
Chemistry
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

34 g/100 mL

Explanation:

The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.

The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:

S = 0.17 kg/5L

S = 170g/5000mL

S = 170g/(5*1000)mL

S = 34 g/100 mL

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When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:
posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

<u>Answer:</u> The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

  • <u>For copper metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = \frac{1}{2}\times 6=3mol of copper metal

Moles of copper reacted = 3 moles

  • <u>For copper(II) nitrate:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = \frac{1}{2}\times 6=3mol of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

  • <u>For silver metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = \frac{2}{2}\times 6=6mol of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

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