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3 years ago

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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2 years ago

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3 years ago

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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.

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

Explanation:

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

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

For sodium hydroxide:

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

In monoprotic acid:

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 =

ValentinkaMS [17]

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,

Concentration of $H^+$ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.090)$

$pH=1.05$

The pH of the solution is, 1.05

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95$

The pOH of the solution is, 12.95

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12.95=-\log [OH^-]$

$[OH^-]=1.12\times 10^{-13}M$

The $OH^-$ concentration is, $1.12\times 10^{-13}M$

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3 years ago

Calculate the mass of 5mole of sodium​

Calculate the mass of 5mole of sodium