Ten name if this compound is Potassium Oxide
Assuming this is a true or false question, the answer would be True
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
Answer:
A. The reaction will proceed forward forming more CH4
B. The reaction will proceed forward forming more CH4
C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.
D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4
E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.
F. The following will favour CH4 at equilibrium
i. Catalyst to the reaction mixture,
ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture
iii. Adding more C to the reaction mixture.