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ELEN [110]
3 years ago
9

The standard enthalpy of formation (ΔHf°) of calcium carbonate is –1207 kJ/mol. Which ONE of the equations below has ΔH° = –1207

kJ? A) Ca(s) + C(s) + 3/2 O2(g) –––––> CaCO3(s) B) 2 Ca(s) + 2 C(s) + 3 O2(g) –––––> 2 CaCO3(s) C) Ca(s) + C(s) + 3 O(g) –––––> CaCO3(s) D) CaCO3(s) ––––> Ca(s) + C(s) + 3/2 O2(g) E) CaO(s) + CO2(g) –––––> CaCO3(s)
Chemistry
1 answer:
Art [367]3 years ago
3 0

Answer:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

Explanation:

Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.

The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:

Ca(s), C(s) and O₂(g)

That means, the equation that represents standard enthalpy of CaCO₃ is:

<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />

<em>Is the equation that has ΔH° = -1207kJ/mol</em>

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