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:
toilet seat and washing of hands ✋ very well
Answer: Outward
Explanation: :The direction of electric field is radially outward for a positive charge and radially inward for a negative charge. Thus, for the electric field points toward south at a position directly south of a positive charge.
Answer:
3.657 kg
Explanation:
Given:
Enthalpy of combustion of hard coal = -35 kJ/g
Enthalpy of combustion of gasoline = 1.28 × 10⁵ kJ/gal
Density of gasoline = 0.692 g/mL
now,
The heat provide 1 gallon of gasoline provide = 1.28 × 10⁵ kJ
and,
heat provided by the 1 gram of coal = 35 kJ
or
1 kJ of heat is provided by (1/35) gram of hard coal
therefore,
For 1.28 × 10⁵ kJ of heat, mass of hard coal = 1.28 × 10⁵ kJ × (1 / 35)
or
For 1.28 × 10⁵ kJ of heat, mass of hard coal = 3657.14 grams = 3.657 kg
