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aksik [14]
3 years ago
14

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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A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
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A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

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brainly.com/question/12942138

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