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2 years ago

6. A 25.0-mL sample of potassium chloride solution was found to have a mass of 25.225 g.

Chemistry

1 answer:

Morgarella [4.7K]2 years ago

4 0

Answer:

6. a. 5.53 (m/m) %; b. 0.7490M

7. 0.156M

Explanation:

6. In the solution of KCl, there are 1.396g of KCl in 25.225g of solution. As mass/mass percent is:

Mass solute / Mass solution * 100

The mass/mass percent of KCl is:

a. 1.396g KCl / 25.225g solution * 100

5.53 (m/m) %

b. Molarity is moles of solute per liters of solution:

Moles KCl -Molar mass: 74.55g/mol-:

1.396g KCl * (1mol / 74.55g) = 0.018726 moles KCl

Volume in liters: 25mL = 0.025L

Molarity:

0.018726 moles KCl / 0.025L = 0.7490M

7. 0.90% means 0.90g of NaCl in 100g of solution:

Moles NaCl -Molar mass: 58.44g/mol-:

0.90g NaCl * (1mol / 58.44g) = 0.0154 moles NaCl

Volume in Liters:

100g * (1mL / 1.01g) = 99mL = 0.099L

Molarity:

0.0154 moles NaCl / 0.099L =

0.156M

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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

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3 years ago

How many molecules are in 42.3g sample of water

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Answer:

The number of molecules is 1.4140*10^24 molecules

Explanation:

To know the number of molecules, we need to determine how many moles of water we have, water has molar mass of 18.015g/mol

This means that one mole of water molecules has a mass of 18.015g.

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2.3480 H2O * 6.022*10^ 23moles/ 1mole of H2O

That's 2.3480 multiplied by 6.022*10^23 divided by 1 mole of H2O

Number of molecules = 1.4140 *10^24 molecules

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Answer:

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