Answer:
The pH is 11
Explanation:
Because pH + pOH = 14
and pOH = - log [OH⁻] = - log (1 x 10⁻³) = 3
we can now calculate pH by manipulating equation one above
pH = 14 - pOH = 14 - 3 = 11
Thus the pH of the solution is 11
if your serious about this question then it is 5
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
Answer:
11419 J/g/ 11.419 KJ/g
Explanation:
H=MCQ
H=225×2.03×(-15-10)
H=225×2.03(25) Note; negative sign is of no use
H=11419J/g
