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Andreas93 [3]
3 years ago
14

What volume (in liters) of a solution contains 0.14 mol of KCl?

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer:

\boxed {\boxed {\sf 0.078 \ L }}

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

  • moles of solute = 0.14 mol KCl
  • molarity= 1.8 mol KCl/ L
  • liters of solution=x

Substitute these values/variables into the formula.

1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}

1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl

1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

The units of moles of potassium chloride cancel.

x= \frac{0.14 }{1.8 L}

x=0.07777777778 \ L

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

x \approx 0.078 \ L

There are approximately <u>0.078 liters of solution.</u>

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Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43
8 0
2 years ago
For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has
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Answer:

70 mL of 5% HCl and 30 mL of 15% HCl

Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

y = 1 - x

This expression is substituted into the first equation and we solve for x.

5x + 15(1 - x) = 8

5x+ 15 - 15x = 8

-10x = -7

x = 7/10 = 0.7

We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

Similarly, the volume of 15% HCl we need is:

(100 mL)(0.3) = 30 mL

3 0
3 years ago
Which two types of information are written in an element's box in the periodic table?
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Yes it is B,D.

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8 0
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Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
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<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

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\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

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Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

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To know more about, solubility product constant, click here,

brainly.com/question/6960236

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