Answer:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)
Answer:
Basically, the more hot the state is, the more kinetic energy it will have. This means that answer D. would be right, as it goes from coldest to hottest states!
Answer:
CH4 +2 O2 — CO2 +2 H2O
Now we see that for 1 mol i.e. 16 grams of methane results in 1 mol of CO2 or 44 grams of CO2.
That means for 3 moles of methane , we will obtain 3 moles of CO2 OR for 48 (3*16) grams of CO2 , we will obtain (44*3) 132 grams of CO2 .That's it….
Explanation:
hey .dude hope the answer was helpful ....
Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
