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3 years ago

How many moles of salt are in 13.8 g of sodium chloride?

Chemistry

2 answers:

ale4655 [162]3 years ago

6 0

Answer: 0.236 mol ≈ 0.24mol

Explanation:

to find the amount of moles of salt in 13.8g of sodium chloride, NaCl, we need to find the mass of sodium in g/mol first.

To find the amount of sodium in g/mol, we sum up the individual masses of

sodium and chloride.

Mass of one mole of sodium = 23g

Mass of one mole of chlorine = 35.4g

Mass of sodium chloride = ( Mass of Na + Mass of Cl)

mass of NaCl = (23 + 35.4) = 58.4g/mol

∴ 1 mole of NaCl = 58.4g

x moles of NaCl = 13.8g

By cross multiplication, we have

13.8g × 1 mol ÷ 58.4g

=0.236mol ≈ 0.24mol

Hoochie [10]3 years ago

3 0

Answer: 0.24 moles

Explanation:

Molecular Mass of NaCl (23 + 35.5) = 58.5g

58.5g of Sodium Chloride -------> 1 mole of NaCl

∴ 13.8g of Sodium Chloride ------> 1 ÷58.5 x 13.8 = 0.2358974 ≈ 0.24moles

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3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

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Explanation :

First we have to calculate the heat gained by the calorimeter.

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c = specific heat = $5.20kJ/^oC$

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$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

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Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

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$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

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T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

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$\Delta U=-2805.8kJ/mol$

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Firdavs [7]

Answer:

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Explanation:

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Question 18 (Essay Worth 8 points)

DENIUS [597]

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learn more:

brainly.com/question/6837993

brainly.com/question/8912651

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