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3 years ago

How many grams of naoh are needed to prepare 500 ml of 0.125 m naoh?

Chemistry

1 answer:

Romashka-Z-Leto [24]3 years ago

6 0

First let us calculate for the number of moles needed:

moles NaOH = 0.125 M * 0.500 L = 0.0625 mol

The molar mass of NaOH is 40 g/mol, hence the mass is:

mass NaOH = 0.0625 mol * 40 g/mol

<span>mass NaOH = 2.5 grams</span>

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Which essential amino acid helps to alleviate insomnia and can be found in turkey?

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Answer:

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4 years ago

Read 2 more answers

An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine

nekit [7.7K]

Answer:

The molecular formula of the compound is $C_{7}H_{6}O_{2}$ .

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

$m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{C} = 84.067\,g$

Hydrogen

$m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{H} = 6.057\,g$

Oxygen

$m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{O} = 31.995\,g$

Now, the number of moles ( $n$ ), measured in moles, of each element are calculated by the following expression:

$n = \frac{m}{M}$

Where:

$m$ - Mass of the element, measured in grams.

$M$ - Molar mass of the element, measured in grams per mol.

Carbon ( $m_{C} = 84.067\,g$ , $M_{C} = 12.011\,\frac{g}{mol}$ )

$n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }$

$n = 7$

Hydrogen ( $m_{H} = 6.057\,g$ , $M_{H} = 1.008\,\frac{g}{mol}$ )

$n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }$

$n = 6$

Oxygen ( $m_{O} = 31.995\,g$ , $M_{O} = 15.999\,\frac{g}{mol}$ )

$n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }$

$n = 2$

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

$C_{7}H_{6}O_{2}$

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3 years ago

Calculate the enthalpy change for the thermite reaction: 2Al(s)+Fe2O3(s)→2Fe(s)+Al2O3(s), ΔH∘rxn=−850 kJ when 12.0 mol of Al und

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<u>Answer:</u> The enthalpy of the reaction for given amount of aluminium will be $-51.0\times 10^2kJ$