Unimelb Chem10003 Question Thread

[Rishi97]

Decided that first year chem students need a thread!

I'll start off:
For the equilibrium CO(g) + H2O(g) --> CO2(g) + H2(g), Kc at 700K is 5.1.
If m mol of all four species are mixed in a vessel of fixed volume at 700K and the system is allowed
to equilibrate, which of the following is true?
A [CO]/[CO2] = 5.1
B [CO]/[CO2] = 2.26
C [CO]/[CO2] = 0.44
D [CO]/[CO2] = 0.19
E [CO]/[CO2] = 1

Any help appreciated 

[Rod]

Ahh awesome thanks for posting this Rishi . I didnt get this, but this is the furthest I got:

CO/CO2 has reversed the equation of the reaction, so we would inverse the equilibrium constant (1/5.1) and it would become 0.19..

But thats the furthest I got, the answer is 0.44...

[grannysmith]

You have to use a little algebra...


If we've got equal moles of reactant and products, then let #moles=m. So, what's the change in moles at equilibrium?
Let n(CO, H₂O, CO₂, H₂)=m. We'll let the change in moles be 'x'.

Because the loss of the reactants is gained by the products.


Taking the inverse,

[Rod]

You have to use a little algebra...


If we've got equal moles of reactant and products, then let #moles=m. So, what's the change in moles at equilibrium?
Let n(CO, H₂O, CO₂, H₂)=m. We'll let the change in moles be 'x'.

Because the loss of the reactants is gained by the products.


Taking the inverse,

Ahh sweet as, thank you so much!!!

[Rishi97]

Indium has two isotopes, 113In and 115In, and an atomic mass of 114.8 amu. If the isotopic
mass of 113In is 112.904 amu and has a natural abundance is 4.29%, the mass of 115In is
closest to:
A. 114.9 amu
B. 115.2 amu
C. 114.7 amu
D. 115.0 amu
E. 115.1 amu

Thanks in advance

[Kel9901]

Indium has two isotopes, 113In and 115In, and an atomic mass of 114.8 amu. If the isotopic
mass of 113In is 112.904 amu and has a natural abundance is 4.29%, the mass of 115In is
closest to:
A. 114.9 amu
B. 115.2 amu
C. 114.7 amu
D. 115.0 amu
E. 115.1 amu

Thanks in advance

Let x be the mass of 115In.
AM=114.8=112.904*4.29/100 + x*(100-4.29)/100=4.8435816+0.9571x
0.9571x=109.9564184
x=114.88, which is closest to A

[Rishi97]

The barium meal used in radiological examinations consists of a suspension of BaSO4 made up in a slurry of Na2SO4. Given that Ksp for BaSO4 is 1.0 x 10-10 calculate the molar solubility of BaSO4 in 0.5M Na2SO4. The answer is:
a) 1.0 x 10-10 M
b) 1.0 x 10-5 M
c) 0.5 x 10-10 M
d) 2 x 10-10 M

I don't even know where to start lol
thnx

[Kel9901]

The barium meal used in radiological examinations consists of a suspension of BaSO4 made up in a slurry of Na2SO4. Given that Ksp for BaSO4 is 1.0 x 10-10 calculate the molar solubility of BaSO4 in 0.5M Na2SO4. The answer is:
a) 1.0 x 10-10 M
b) 1.0 x 10-5 M
c) 0.5 x 10-10 M
d) 2 x 10-10 M

I don't even know where to start lol
thnx

Ksp=1.0*10^-10=c(Ba²⁺)*c(SO₄^2-)

c(SO₄^2-) is approximately 0.5 M (as very little BaSO₄ will dissolve).

Hence c(Ba²⁺)=1.0*10^-10/0.5=2.0*10^-10 M. Since all of the Ba²⁺ will come from the BaSO₄, the molar solubility of the barium sulfate is 2.0*10^-10 M, and therefore D

[TrebleClef]

In the crystal structure of CsCl, the chloride ions adopt a simple primitive packing arrangement and the Cs+ ions occupy the cubic interstitial sites. The cubic unit cell edge length is 412pm. Assuming that the chloride ions just touch each other in this arrangement, what is the radius of the Cs+ cation?

A) 357pm
B)198pm
C)206pm
D)302pm
E)151pm

Thanks in advance!

[Rishi97]

In the crystal structure of CsCl, the chloride ions adopt a simple primitive packing arrangement and the Cs+ ions occupy the cubic interstitial sites. The cubic unit cell edge length is 412pm. Assuming that the chloride ions just touch each other in this arrangement, what is the radius of the Cs+ cation?

A) 357pm
B)198pm
C)206pm
D)302pm
E)151pm

Thanks in advance!

Was there more info to this question? I've done this question but can't remember which exam it was from.

[TrebleClef]

Was there more info to this question? I've done this question but can't remember which exam it was from.

No there wasn't. I've worked this one out, but I'm still really struggling to draw out the structures, and being able to visualize the cube/unit cells. I can't figure out where the cubic/tetrahedral/octahedral sites and stuff are. I'm just in way over my head. I can do the calculations, but I'm just having so much trouble visualizing them.