Answer:
The partial pressure of neon in the vessel was 239 torr.
Explanation:
In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.
Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:
PT= P1 + P2 + P3 + P4…+ Pn
where n is the amount of gases present in the mixture.
In this case:
PT=PN₂ + PAr + PHe + PNe
where:
- PT= 987 torr
- PN₂= 44 torr
- PAr= 486 torr
- PHe= 218 torr
- PNe= ?
Replacing:
987 torr= 44 torr + 486 torr + 218 torr + PNe
Solving:
987 torr= 748 torr + PNe
PNe= 987 torr - 748 torr
PNe= 239 torr
<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>
Formal charge = valence electron - bonds - dots.
double bonded oxygen = 6-2-
4 so it is 0
single bonded oxygen = 6-1-6 so it is -1
sulfur = 6-3-2 so it is +1
thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
Answer:
The answer to your question is 50 moles of O₂
Explanation:
Balanced Chemical reactions
1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)
moles of N₂(g) = 20 moles
moles of O₂(g) = ?
Process
1.- Calculate the moles of NH₃
1 mol of N₂ ------------- 2 moles of NH₃
20 moles of N₂ --------- x
x = (20 x 2) / 1
x = 40 moles of NH₃
2.- Calculate the moles of O₂
4 moles of NH₃ -------------- 5 O₂
40 moles of NH₃ ------------ x
x = (40 x 5) / 4
x = 200 / 4
x = 50 moles of O₂
