Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Answer:
2.2×10^8
Explanation:
Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20
2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14
Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)
Overall ionic reaction:
Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)
Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2
Keq= 2.2×10^8
Kw= ion dissociation constant of water
Boyle’s = increase as volume decreases
Charles = increases and pressure increases
Gay-lussacs = increases as pressure increases
Answer:
It is the closest sir.
Have an amazing day and enjoy. P.S Good luck!
Explanation:
Answer:
PV=nRT
Explanation:
V=<u>R</u><u>T</u><u>n</u>
P
rearrangement gives
nT
where P=pressure
V=volume
n=number of moles
R=ideal gas(0.0820atmdm/3 mol/k)
T=temperature in kelvin
