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MAXImum [283]
3 years ago
13

Sea water contains roughly 28.0 g of nacl per liter. What is the molarity of sodium chloride in sea water?

Chemistry
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

The molarity of sodium chloride in sea water is 0.479 M

Explanation:

Step 1: Data given

Mass of NaCl = 28.0 grams

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 28.0 grams / 58.44 g/mol

Moles NaCl = 0.479 moles

Step 3: Calculate molarity NaCl in sea water

Molarity = moles / volume

Molarity NaCl = 0.479 moles / 1L

Molarity of NaCl in sea water = 0.479 mol/L = 0.479 M

The molarity of sodium chloride in sea water is 0.479 M

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel
n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557

<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 nitrogen gas:</u>

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol

  • <u>For ammonia gas:</u>

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B+n_C}

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

\chi_{(H_2)}=\frac{1.025}{1.84}=0.557

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

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3 years ago
24. All elements found on the left side of the Periodic Table of
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Answer:

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Explanation:

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Answer:

0.0745 mole of hydrogen gas

Explanation:

Given parameters:

Number of H₂SO₄ = 0.0745 moles

Number of moles of Li = 1.5107 moles

Unknown:

Number of moles of H₂ produced = ?

Solution:

To solve this problem, we have to work from the known specie to the unknown one.

The known specie in this expression is the sulfuric acid,  H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.

   Balanced chemical equation:

                    2Li   +     H₂SO₄   →   Li₂SO₄   +   H₂

 From the balanced equation;

     

Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.

           2 mole of Li reacted with 1 mole of  H₂SO₄

          1.5107 mole of lithium will react with \frac{1.5107}{2}  = 0.7554mole of H₂SO₄

But we were given 0.0745 moles,

This suggests that the limiting reagent is the sulfuric acid because it is in short supply;

   

   since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;

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