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zmey [24]
3 years ago
6

How many of each kind of atom is in one molecule of water? 1 hydrogen atom and 2 oxygen atoms 1 hydrogen atom and 1 oxygen atom

2 hydrogen atoms and 1 oxygen atom 2 hydrogen atoms and 2 oxygen atoms
Chemistry
1 answer:
umka21 [38]3 years ago
7 0

Answer:

What period is the atom in?

1.

What group is the atom in?

What is the name of this atom?

What other atom would have similar

properties to the atom?

Will this atom conduct electricity well

Explanation:

How many of each kind of atom is in one molecule of water? 1 hydrogen atom and 2 oxygen atoms 1 hydrogen atom and 1 oxygen atom 2 hydrogen atoms and 1 oxygen atom 2 hydrogen

and 2 oxygen atoms

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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 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. 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_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

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

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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