Answered

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Consider the intermediate chemical reactions:

[tex]\[
\begin{array}{ll}
Ca(s) + CO_2(g) + \frac{1}{2}O_2(g) \rightarrow CaCO_3(s) & \Delta H_1 = -812.8 \, \text{kJ} \\
2Ca(s) + O_2(g) \rightarrow 2CaO(s) & \Delta H_2 = -1,269.8 \, \text{kJ}
\end{array}
\][/tex]

The final overall chemical equation is:

[tex]\[
CaO(s) + CO_2(g) \rightarrow CaCO_3(s)
\][/tex]

When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation:

A. is halved and has its sign changed.
B. is halved.
C. has its sign changed.
D. is unchanged.

Sagot :

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