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>
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Answer:2.86x10^-7m
Explanation:E=hc/^
E=6.94x10^-19J
c = 2.9979x10^8m/s
h= 6.626x10^-34Js
^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19
= 2.86x10^-7m
Answer:
0.185moles of Al₂O₃
Explanation:
Mass of Al = 10g
Mass of O₂ = 19g
Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃
This is the balanced reaction equation.
Solution
From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.
Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.
Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:
Number of moles of Al = 
= 
= 0.37mol
From the equation:
4 moles of Al produced 2 moles of Al₂O₃
0.37 mole will yield: 
= 0.185moles of Al₂O₃
V1/T1 = V2/T2
24.5 + 273 = 298.5K
46.9 + 273 = 319.9K
95.8/298 = V2/319.9
V2 = 102.66L
