Please i need the answer for 18,19,20Answer and I will give you brainiliest

Chemistry

1 answer:

dezoksy [38]3 years ago

8 0

Number 18 is correct, number 19 is the particles speed up and the substance expands, and number 20 is decrease in volume

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Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

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

Explanation:

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:

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