First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:
pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M
M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7
pH = -log[<span>1 x 10^-7</span>]
pH = 7
Explanation:
an increase in concentration increases the rate of the reaction. This is because there are more reactant particles available which allows for more effective collisions between reactant particles in a given period of time. More effective collisions bring about a faster rate of reaction.
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:
a) 25%
b) 27.5 g
c) 90%
Explanation:
a) 75% fat-free by weight means 25% of the weight is made by fat.
b) 110 g ___ 100%
x ___ 25%
x = 27.5g
Each hot dog has 27.5g of fat.
c) 9 cal ___ 1 g fat
y ___ 27.5 g fat
y = 247.5 cal
275 cal ___ 100%
247.5 cal ___ z
z = 90%
90 % of the calories come from fat.
