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Orlov [11]
3 years ago
11

How many molecules of HCO3 are in 4 moles of the substance?

Chemistry
1 answer:
never [62]3 years ago
8 0
This question is about Avogadro's number.


Avogadro's number is the number is 6.022 * 10^23.


That means that 1 mol of molecules is 6.022 * 10^23  molecules, so 4 moles is 4 * 6.022 * 10^23 molecules = 24.088 * 10^23 molecules = 2.4088 * 10^24 molecules.



Answer: 2.4088 * 10^24 molecules.
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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

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

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

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

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

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Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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