![\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}](https://tex.z-dn.net/?f=%5Chuge%7B%5Ctextbf%7B%5Ctextsf%7B%7B%5Ccolor%7Bpink%7D%7BAn%7D%7D%7B%5Cred%7Bsw%7D%7D%7B%5Corange%7Ber%7D%7D%20%7B%5Ccolor%7Byellow%7D%7B%3A%7D%7D%7D%7D%7D)
As depth increases, the density of the layers decreases.
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Should be , b
. Not positive tho
No a molecule is 2 different atoms bond