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3 years ago

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What are the 4 factors that affect the dissolving of a gas

1 answer:

lara [203]3 years ago

7 0

Nature of the solute and solvent - The amount of solute that dissolves depends on what type of solute it is. While only 1 gram of lead (II) chloride can be dissolved in 100 grams of water at room temperature, 200 grams of zinc chloride can be dissolved. This means that a greater amount of zinc chloride can be dissolved in the same amount of water than lead II chloride.
Temperature - Increasing the temperature not only increases the amount of solute that will dissolve but also increases the rate at which the solute will dissolve. For gases, the reverse is true. An increase in temperature decreases both solubility and rate of solution.

Pressure - Changes in pressure have practically no effect on solubility. For gaseous solutes, an increase in pressure increases solubility and a decrease in pressure decreases solubility. Example: When the cap on a bottle of soda pop is removed, pressure is released, and the gaseous solute bubbles out of solution. This escape of a gas from solution is called effervescence.

Stirring - Stirring brings fresh portions of the solvent in contact with the solute. Stirring, therefore, allows the solute to dissolve faster.

Hope this helps!! (If not I'm sorry!)

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Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

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3 years ago

According to the following balanced chemical equation, if you want to produce eight moles of H2O, how many moles of CH4 do you n

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2 years ago

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500 mL of water is added to 400 mL of 0.35 M HCl. Find the concentration of the diluted solution.

jarptica [38.1K]

The concentration of diluted solution is 0.16 M

<u>Explanation:</u>

As, the number of moles of diluted solution and concentrated solution will be same.

So, the equation used to calculate concentration will be:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated HCl solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted HCl solution

We are given:

$M_1=0.35M\\V_1=400mL\\M_2=?M\\V_2=(500+400)mL=900mL$

Putting values in above equation, we get:

$0.35\times 400=M_2\times 900\\\\M_2=\frac{0.35\times 400}{900}=0.16M$

Hence, the concentration of diluted solution is 0.16 M

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