Hi people, its me again :) Here are some chem mcq questions that i cant solv :(e even though i feel there is nothing wrong with my method. my workings will be posted so you all can correct my mistakes. thanks for your time :)
16 Steam, at an initial pressure of 1 atm, dissociates at T K according to the following equation: 2H2O(g) <---> 2H2(g) + O2(g)
If the total pressure at equilibrium is 1.3 atm, what is the numerical value of the equilibrium constant, Kp of the reaction at T K? ans: D
A 0.108
B 0.270
C 0.450
D 0.675
I started off with my ICE table (let equilibrium partial pressure of H2 be x): I 1 0 0
C -x +x + x/2
E 1-x x x/2
since total pressure at equilibrium is 1.3, (1-x) + x + (x/2) = 1.3
--> x = 0.6 which means that at equilibrium, the pressure of H20, H2 and O2 is 1.4, 0.6 and 0.3 respectively.
so Kp = (0.3)(0.6)^2/(1.4)^2 = 0.055
Damn, for some reason i cant edit my lines. The wall of text wasnt intended. Can the mods help? thanks.
Solved
Which of the following sample of gas contains the same number of atoms as 1 g of hydrogen gas?
A 6 g of steam
B 8 g of methane gas
C 22 g of carbon dioxide
D 32 g of oxygen gas
amount of H2 in 1g = 1/2 = 0.5mol
for A), amount of H20 in 6g = 6/18 = 1/3mol
for B), amount of CH4 in 8g = 8/16 = 1/2mol
for C), amount of C02 in 22g = 22/44 = 1/2mol
for D), amount of O2 in 32g = 32/(16x2) = 1
so how can it the answer be option A? I think i'm making a very noob mistake but i cant see it now. someone please enlighten me.
10 cm3 of a gaseous hydrocarbon is completely burnt in 40 cm3 of oxygen. After cooling to room temperature, the residual gas is passed through aqueous KOH,
The volume of gas decreases to 15 cm3. What is the molecular formula of the hydrocarbon?
A CH4
B C2H2
C C2H6
D C3H6
ans: B
so as always: CxHy + (x+y/4)O2 ---> xCo2 + (y/2)H20
10cm3 15cm3 25cm3
my volume of Co2 is 25cm3 as the question said that "After cooling to room temperature, the residual gas is passed through aqueous KOH, The volume of gas decreases to 15 cm3. " So 40cm3 - 15cm3(vol of remaining oxygen) = volume of C02 = 25cm3
Is this step right?
If it is, then by mol ratio, 25/10 =x/1 ; x = 5/2 (but how can x be a fraction?)
then using mol ratio of O2 = CO2, y would be 4.
Originally posted by Lord dejavu:Which of the following sample of gas contains the same number of atoms as 1 g of hydrogen gas?
A 6 g of steam
B 8 g of methane gas
C 22 g of carbon dioxide
D 32 g of oxygen gasamount of H2 in 1g = 1/2 = 0.5mol
for A), amount of H20 in 6g = 6/18 = 1/3mol
for B), amount of CH4 in 8g = 8/16 = 1/2mol
for C), amount of C02 in 22g = 22/44 = 1/2mol
for D), amount of O2 in 32g = 32/(16x2) = 1so how can it the answer be option A? I think i'm making a very noob mistake but i cant see it now. someone please enlighten me.
Each molecule of H2 gas contains 2 atoms. Remember now we are taking about atoms and not molecules.
0.5 mole of H2 gas contains 0.5 X 2 X 6.023X10^23 atoms not 0.5 X 6.023X10^23
Hopefully you understand this now. Alot of students forgot that most gas molecules contains more than 1 atoms per molecule.
20.0 cm3 of 0.02 mol dm–3 aqueous sodium bromate(V), NaBrO3, is found to react completely with 80.0 cm3 of 0.01 mol dm–3 hydroxylamine, NH2OH. The half equation for the reduction of bromate(V) ion is given as shown.
Mole ratio=BrO3- : NH2+ is 1:2
BrO3–(aq) + 6H+(aq) + 6e—> Br– (aq) + 3H2O(l)
The eqn tells u that 1 mole of Bro3- requires 6e.
so 1 mole of nh2+ releases 3e-.
Since its oxidation no. in nh2+ is -1, when it releases 3 e- its new oxidation will be +2.
therefore the answers is NO. (option b)
Thanks dkcx and arxoz, really appreciated.