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goblinko [34]
3 years ago
10

A lab technician mixes a 0.730 M solution of sodium bromide (NaBr) and water. The volume of the solution is 135 milliliters.

Chemistry
2 answers:
GarryVolchara [31]3 years ago
6 0

Answer: 10.1

Explanation:

Ivenika [448]3 years ago
3 0

Answer:

We need 10.14 grams of sodium bromide to make a 0.730 M solution

Explanation:

Step 1: Data given

Molarity of the sodium bromide (NaBr) = 0.730 M

Volume of the sodium bromide solution = 135 mL = 0.135 L

Molar mass sodium bromide (NaBr) = 102.89 g/mol

Step 2: Calculate moles NaBr

Moles NaBr = Molarity NaBr * volume NaBr

Moles NaBr = 0.730 M * 0.135 L

Moles NaBr = 0.09855 moles

Step 3: Calculate mass of NaBr

Mass NaBr = 0.09855 moles * 102.89 g/mol

Mass NaBr = 10.14 grams

We need 10.14 grams of sodium bromide to make a 0.730 M solution

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The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.

Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.

Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.

The question is incomplete. Please read below to find the missing content.

Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.

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