Using this information together with the standard enthalpies of formation of o2(g), co2(g), and h2o(l) from appendix c, calculate the standard enthalpy of formation of acetone.
complete combustion of 1 mol of acetone (c3h6o) liberates 1790 kj:
c3h6o(l)+4o2(g)? 3co2(g)+3h2o(l)? h? =? 1790kj

ΔHacetone = - 247.5 kJ/mol

Explanation:

The enthalpy equation is as follows

ΣnΔHproducts – ΣmΔHreactants =ΔHreaction

3×ΣnΔH(CO2(g)) + 3×ΣnΔH(H2O) - ΣnΔH (C3H6O(l)) =ΔHreaction = -1790 kJ/mol

[3(-393.5) + 3(-285.8)] – ΔHacetone

= -1790 kJ/mol

(-1180.5 – 857.4)kJ/mol - ΔHacetone =

-1790 kJ/mol

-2037.9 kJ/mol - ΔHacetone

= -1790kJ/mol

-2037.9 kJ/mol + 1790kJ/mol = ΔHacetone

- 247.5 kJ/mol = ΔHacetone

ΔHacetone = - 247.5 kJ/mol

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms