Molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
The geometry around left carbon that is CH₃ is tetrahedral.
As the hybridization around left carbon is sp³ that shows its geometry should be tetrahedral and as there are 4 ligands around carbon and there is no lone pair present so the geometry is tetrahedral. So, the molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Answer:
a minimum of <em>1</em><em>0</em><em>,</em><em>0</em><em>0</em><em>0</em><em> </em>years
Explanation:
a) 4P + 3O2 --> 2P2O3
b) The chemical reaction above limits the number of molecules of P2O3 produced for every 4 atoms of P.
