The Germans were the first to send up a satellite into space.
2 answers:
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First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
(A) > (B) > (c)
B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C
D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules
∵ Mw A < Mw B < Mw C
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C
e) the rank of density:
when ρ = m/ v
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C
F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C
∴the rank of the collision frequency is:
A > B > C
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
(A) > (B) > (c)
B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C
D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules
∵ Mw A < Mw B < Mw C
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C
e) the rank of density:
when ρ = m/ v
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C
F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C
∴the rank of the collision frequency is:
A > B > C
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M