What is the equilibrium constant, Kc, if [A] = 0.19 M, [B] = 0.952 M, and [C] =

Question

What is the equilibrium constant, Kc, if [A] = 0.19 M, [B] = 0.952 M, and [C] = 0.70 M 2 A ReverseEquilibrium 2 B + C 18 3.3 3.5 7.0 Which change will increase the quantity of product in the following reaction? 2 SO_3(g) ReverseEquilibrium 2 SO_2(g) + O_2(g) Delta H = 198 kJ an decrease in temperature an increase in volume of adding a catalyst a increase in pressure Arrhenius base is defined as a proton donor proton acceptor that produces hydrogen ions in water that produces OH^- ion In the reaction H_2CO_3 + H_2O ReverseEquilibrium HCO_3^- + H_3O^+, the Bronsted acid H_2CO_3 and H_2O HCO_3^- and H_2CO_3 H_2O and H_3O^+ H_3O^+ and H_2CO_3

Explanation / Answer

Q16

Kc for:

2A <->2B + C

K = [products]^p/[reactnats]^r

K = [B]^2[C]/ [A]^2

K = (0.952^2)(0.70) / (0.19^2) = 17.573

nearest answer is a, k = 18

Q17

increase in produt...

a) false, decreasing T will not favour this, since endothermic

b) incresae in volume decreases Pressure, this favours most mol formation, so it is TRUE

c) catalyst won't change equilibirum, only rates

d) increas in P, favours the least mol formation, as stated in b, false

Q18

Ahrrenius base is

a) this is an acid

b) this is a bronsted lowry acid

c) Ahrrenius acid... by definition

d) this is a Ahrrenius base!, choose it

Q19

bronsted acid --> must donate H+

so..

H2CO3 is donating H+ to form HCO3-

this is the acid then H2CO3

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