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1. does preparing food in different ways affect the calories and percents of certain things
2. prepare 3 foods in different types of ways(bake, fry, grill, microwave, boil, etc.)
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
The equilibrium constant is found by [product]/[reactant]
If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.
At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.
At equilibrium, for your reaction, it is predominantly reactants.
Answer:
90 m
Explanation:
60 m plus 30 m=90 m simple math