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Viefleur [7K]
3 years ago
8

Calculate the mass of 5mole of sodium​

Chemistry
1 answer:
GREYUIT [131]3 years ago
4 0

Answer: 115 g Na

Explanation: solution attached:

Convert moles of Na to grams using its molar mass.

5 moles Na x 23 g Na / 1 mole Na

= 115 g Na

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Which type of monomer combines and forms polypeptides?
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By titration, 15.0 mLmL of 0.1008 MM sodium hydroxide is needed to neutralize a 0.2053-gg sample of an organic acid. What is the
andre [41]

The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

<u>Answer:</u> The molar mass of monoprotic acid is 135.96 g/mol

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For sodium hydroxide:</u>

Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of OH^- ion of NaOH neutralizes 1 mole of H^+ ion of monoprotic acid

So, 0.00151 moles of OH^- ion of NaOH will neutralize \frac{1}{1}\times 0.00151=0.00151mol of H^+ ion of monoprotic acid

<u>In monoprotic acid:</u>

1 mole of H^+ ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of H^+ ion will be released by \frac{1}{1}\times 0.00151=0.00151mol of monoprotic acid

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol

Hence, the molar mass of monoprotic acid is 135.96 g/mol

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3 years ago
Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =
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Answer : The concentration of OH^- ion, pH and pOH of solution is, 1.12\times 10^{-13}M, 1.05 and 12.95 respectively.

Explanation : Given,

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12.95=-\log [OH^-]

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