(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M
Carbon dioxide can’t exist in three states; Gas, Liquid & Solid. At normal temperatures and pressures, CO2 is colorless with a slightly pungent odor at high concentrations. If compressed and cooled to proper temperature the gas liquifies. Solid CO2, (dry ice) sublimates back to the natural gaseous state.
I believe that would be Cadmium Permaganate
Hope that helped :)
It would go B. A. E. D. C.
Hope I helped!
Plant cells are affected as they begin to wilt and reduce photosynthesis
