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3 years ago

134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.

Chemistry

2 answers:

lianna [129]3 years ago

7 0

Answer:

0.004 m

Explanation:

Given data:

Mass of nitric acid = 132 g

Mass of water = 512 g

Molality of nitric acid = ?

Solution:

Formula:

Molality = number of moles of solute / mass of solvent in Kg

Number of moles of nitric acid:

Number of moles = mass/molar mass

Number of moles = 132 g/ 63.01 g/mol

Number of moles = 2.09 mol

Molality:

m = 2.09 mol / 512 Kg

m = 0.004 m

Nana76 [90]3 years ago

6 0

Answer: 4.15234 m

512 g H2O * $\frac{1 kg}{1000 g}$ = 0.512 kg H2O

Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ * $\frac{mol}{63.012 g}$ = 2.126 mol

m = $\frac{2.126 mol}{0.512 kg}$ = 4.15234 m

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<h3>Further explanation</h3>

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Read 2 more answers

A container initially holds 1.24mol of hydrogen gas and has a volume of 27.8L. Hydrogen gas was added to the container, and the

Fantom [35]

Answer:

After increasing the volume, we have 1.81 moles of hydrogen gas in the container

Explanation:

Step 1: Data given

Number of moles hydrogen gas (H2) = 1.24 moles

Volume of hydrogen gas (H2° = 27.8 L

The final volume is increas to 40.6 L

Step 2: Calculate the new number of moles

V1/n1 = V2/n2

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⇒with V2 = the final volume = 40.6 L

⇒with n2 = the new number of moles = TO BE DETERMINED

27.8L / 1.24 moles = 40.6 L / n2

n2 = 40.6 / (27.8/1.24)

n2= 1.81 moles

After increasing the volume, we have 1.81 moles of hydrogen gas in the container

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If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w

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Answer:

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