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:
particles in 2 moles.
Explanation:
The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.
It is also known as Avogadro's constant.
∴ Number of particles in one mole = 
∴ Number of particles in 2 mole = 2 times Number of particles in one mole
∴ Number of particles in 2 mole=
Hence there are particles in 2 moles.
Answer:
I think the teaspoon will cool down faster.
Explanation:
The teaspoon of water will cool down faster because it is a smaller amount of water so the entire [drop] will absorb heat energy faster.
