CSE 2010 May Exam Questions!

[Matt-Ong]

I just finished Section A for the CSE 2010 May exam in 20 minutes, which I am pretty proud of as it was the 4th exam I've done
But I've a few questions that I did not understand :/

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Q2)
9.30g of element M formed an oxide with an empirical formula of MO₃ and a mass of 17.88g. In what section of the periodic table does this element belong to?
A. Transition Metals  <--- Correct answer
B. Alkali Metals
C. Alkaline earths
D. Halogens

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Q6) Consider the 2 molecules below

Molecule I : CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₃
Molecule II : CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂COOH

When seperating components using high performance liquid chromatography (HPLC) with a polar stationary phase, molecule I has the
A. longer retention time because it has a higher molar mass.
B. the shorter retention time because it has a higher molar mass.
C. longer retention time because it is non-polar
D. shorter retention time because it is non-polar <-- Correct Answer

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Q8) The peak at 45 of propan-2-ol is most likely to be (Note: Peak at 45 is the base peak, so does that mean the strongest bond?)
A. CH₃CHOH
B. CH₃CH₂O⁺
C. CH₃CHOH⁺  <--- Correct answer
D. CH₂CH₂OH⁺

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Those are basically all the questions I got wrong.. The answers didn't really help as they just stated what the answer was without any explanation :/

[b^3]

Could you put the answers up so that we can check that our thinking is right? so that we dont explain a wrong answer.

[Water]

6) C

C

2) A

Not sure if I'm right.

[Matt-Ong]

Could you put the answers up so that we can check that our thinking is right? so that we dont explain a wrong answer.

Yeap.. Sorry about that.. Modified the questions

6) C

C

2) A

Not sure if I'm right.

You've got 2 and 8 right! Could you explain to me how'd you figure out the answer?

[b^3]

For 2) i would say A but this is assuming the Empirical formula is the same as the molecular formula, then work out the mass of M in MO₃ and equate it to the original mass and solve, but i may have over complicated it

this would put it in the transition metals and this kinda makes sense for the O₃ since they tend to bond in weird ways.

For 6) since it is non-polar and the stationary phase is polar, the retention time is shorter because it is less attracted to the stationary phase than the other molecule and so will be eluted in a lesser time that the other molecule. The polar molecule will "stick" to the stationary phase more and so take longer to elude.

[Water]

Herpy Derp, Question 6 , lol I misread it.


a)

9.30g = Element M

m(MO3) = 17.88gram

m(O3) = 8.58gram

n(O3) = 8.58gram/16gramol-1
         = 0.53625 mol

n(M)= 1/3 x n(O3)
      = 0.17875 mol

M(Element M) = 9.30gram/0.17875 mol

                   = 52 gram = Chromium


Because CH3 CH2 - OH    CH3


The OH , in here would be most vulnerable to being bombarded by electrons.

I"m using logic similarly to CH3 CH2 - CL CH3 when it is substituted with NaOH.






For question 6. To me, D would be correct as well, I misread question sorry and didn't look at when it was comparing between the two molecules. But, the polarity would cause it to adsorb much strongly to the stationary phase. Whereas the weight would just cause clogging amongst the compounds. Logic and reasoning I guess. >_<!?


Perhaps a university level chemistry student, can enlighten us

[ninbam1k]

are there any solutions for this paper? I couldn't find it :/

[luffy]

I just finished Section A for the CSE 2010 May exam in 20 minutes, which I am pretty proud of as it was the 4th exam I've done
But I've a few questions that I did not understand :/

**********************************
Q2)
9.30g of element M formed an oxide with an empirical formula of MO₃ and a mass of 17.88g. In what section of the periodic table does this element belong to?
A. Transition Metals  <--- Correct answer
B. Alkali Metals
C. Alkaline earths
D. Halogens

***********************************
Q6) Consider the 2 molecules below

Molecule I : CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₃
Molecule II : CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂COOH

When seperating components using high performance liquid chromatography (HPLC) with a polar stationary phase, molecule I has the
A. longer retention time because it has a higher molar mass.
B. the shorter retention time because it has a higher molar mass.
C. longer retention time because it is non-polar
D. shorter retention time because it is non-polar <-- Correct Answer

***********************************

Q8) The peak at 45 of propan-2-ol is most likely to be (Note: Peak at 45 is the base peak, so does that mean the strongest bond?)
A. CH₃CHOH
B. CH₃CH₂O⁺
C. CH₃CHOH⁺  <--- Correct answer
D. CH₂CH₂OH⁺

***********************************

Those are basically all the questions I got wrong.. The answers didn't really help as they just stated what the answer was without any explanation :/

Sorry, I could be wrong in saying this, but:

Q2) Since its , can't we simply conclude that the oxidation state of M is +6. I realise oxidation states do not imply the valency of an ion, but in this case, it does seem like a basic ionic bond. An oxidation no. of +6, means it would have to be a transition metal.

3)Molecule I: Non-Polar
   Molecule II: Polar
Stationary Phase: Polar
Hence, Mobile Phase: Non-polar
- While molecular mass does make a difference if the two objects differ greatly, or if they are both polar or both non-polar, its affect is completely masked by intermolecular forces in this case. Hence, Molecule II would have a longer retention time, due to stronger attraction (adsorption) to the stationary phase.
- Therefore, Molecule I, as asked by the question, would have a shorter retention time, due to weaker attraction to the stationary phase, and stronger attraction to the mobile phase, as it is non-polar.

Propan-2-ol has a molar mass of approximately 60. Hence, for a base peak of 45, a group must have been taken off. Hence, the compound must be .
- I don't think D would be a correct answer. Since its propan-2-ol and a CH3 group was taken off, the Carbon with the -OH group must have only 3 bonds. In this case, it has all 4 bonds and the adjacent carbon has only 3. This would occur in Propan-1-ol, but not Propan-2-ol.
(Draw a diagram of propan-2-ol  - it will make my explanation a lot clearer... I hope)

[SimoHDK]

Another CSE Question:

Question 7
In a gravimetric analysis of salt in pond water, 25mL of the pond water had excess silver nitrate solution added to it. When compared with a volumetric analysis result, the calculated concentration of the salt was less for gravimetric analysis. A possible explanation for this result is that:

a) the precipitate had not been dried to constant mass

b) there were other anions that precipitated as well as the silver chloride

c) some of the precipitate had remained in the reaction flask

d) the silver chloride precipitate in the filter paper had not been washed with distilled water.

[lilaznkev1n]

Another CSE Question:

Question 7
In a gravimetric analysis of salt in pond water, 25mL of the pond water had excess silver nitrate solution added to it. When compared with a volumetric analysis result, the calculated concentration of the salt was less for gravimetric analysis. A possible explanation for this result is that:

a) the precipitate had not been dried to constant mass

b) there were other anions that precipitated as well as the silver chloride

c) some of the precipitate had remained in the reaction flask

d) the silver chloride precipitate in the filter paper had not been washed with distilled water.

I think it's C because if some of the precipitate had remained in the flask, which will result in less mass of precipitate calculated which will lower the concentration.
In A,B,D it will result in a greater mass of precipitate calculated which will increase in concentration of salt in the water.