Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.
