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3 years ago

How many moles in 4.3 x 1024 atoms of sodium (Na)?

Chemistry

1 answer:

wel3 years ago

3 0

Answer:How many ions are in 2 moles of NaCl?

2 mol NaCl = 2 x 6.02 x 10 23 = 1.204 x 10 24 NaCl formula units 2 mol NaCl = 2 (2 x 6.02 x 10 23 )= 2 (1.204 x 10 24) = 2.408 x 10 24 atoms *2 moles of sodium chloride contains 2.408 x 10 24 ions (combination of Na + ions & Cl – ions). H

Explanation:

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<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

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Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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