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3 years ago

Describe how you would prepare a supersaturated solution.

Chemistry

2 answers:

earnstyle [38]3 years ago

5 0

Answer:

To prepare a supersaturated solution, the added amount must be higher than the solubility for the given volume of solvent

Explanation:

Hello,

This could be answered by knowing that all the solutes have a property called solubility which accounts for the maximum amount of it that can be thoroughly dissolved into a specific solvent. Thus, to prepare a supersaturated solution, the added amount must be higher than the solubility for the given volume of solvent at a specific temperature. For example, at 20°C, 45.8g of aluminium chloride are completely dissolved into 100 mL of water, so at that amount, the solution will be saturated, thus, if one adds more than 45.8g the solution will start being supersaturated.

Best regards.

Anastasy [175]3 years ago

4 0

To make a supersaturated solution, make a saturated solution of sugar by adding 360 grams of sugar to 100 mL of water at 80 degrees Celsius. When the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water should only dissolve 210 grams of sugar.

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murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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notka56 [123]

Answer:

Explanation:

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Zarrin [17]

Answer:

Explanation:

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2 years ago

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Determine the volume, in liters, of 3.2 mol of CO2 gas at STP.

Allushta [10]

Answer:

71.7 L

Explanation:

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V = volume (L)

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T = temperature (K)

According to the information provided in this question;

P = 1 atm (STP)

V = ?

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3 years ago

Which of the following is not part of the proper protocol for using acids and bases?

Anni [7]

Answer:

B.Add acid to water,not water to acid

Explanation:

they should not be mixed

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2 years ago