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Serggg [28]
3 years ago
7

Which example is not a structural adaptation of a plant? A. curling around a tree trunk B. protecting itself with thick bark C.

protecting itself with needles D. attracting pollinators with bright colored flowers
Chemistry
1 answer:
aalyn [17]3 years ago
3 0

A. curling around a tree trunk

I believe would be the correct answer because it is not an adaptation that it has gained through evolution and it is something that happens after the plant has grown.

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)
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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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