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Author Topic: Avogadro Chem Contest  (Read 1521 times)
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Computerized
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« on: May 04, 2009, 05:12:45 PM »

I have a number of questions from previous Avogadro contests.


1. At 0C and 101.3 kPa, a 2.75L sample of a particular gas weighs 5.39g. Which gas could it be?

A. N2
*B. N2O
C. NO
D. F2
E. NF3


2. In which one of the following is hexane, C6H14 (l), most soluble?

A. NaCl (aq)
B. HOCH2CH2OH (ethylene glycol)
C. CH3COOH (aq) (vinegar)
*D. C6H6 (benzene)
E. H20 (Water)


For #1, do you use the combined gas laws formula, where VP/T equals some constant? I'm not sure where you go after that though.

For #2, how on Earth are you supposed to know that? Is it just general knowledge, or can you use process of elimination, or is there some other way?

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« Reply #1 on: May 04, 2009, 07:36:15 PM »

2. In which one of the following is hexane, C6H14 (l), most soluble?

A. NaCl (aq)
B. HOCH2CH2OH (ethylene glycol)
C. CH3COOH (aq) (vinegar)
*D. C6H6 (benzene)
E. H20 (Water)

For #2, how on Earth are you supposed to know that? Is it just general knowledge, or can you use process of elimination, or is there some other way?

Think about the forces which exists between C6H14 molecules and then think about the forces between the solvent molecules. Which one(s) is weaker? Which one(s) are stronger? What happens when something dissolves in something else, in terms of the forces of attraction?
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« Reply #2 on: May 04, 2009, 09:10:48 PM »

1. At 0C and 101.3 kPa, a 2.75L sample of a particular gas weighs 5.39g.

How many moles of the gas? What must be its molar mass?

2nd question calls for application of "similar dissolves similar" idea.
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« Reply #3 on: May 05, 2009, 06:40:58 AM »

Ah, I see, so for #1 you're supposed to use the Ideal Gas Law. Had to learn that just now by myself haha.

For #2, I guess you're supposed to realize then that hexane is non-polar, and then by process of elimination you can figure out that benzene is the only other one that's non-polar?

Thanks for the help. I may have more questions in the future.
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« Reply #4 on: May 07, 2009, 05:18:32 PM »

Few more questions I'm having trouble with...

1. An aqueous solution of FeCL3 has a density of 1.19 g/ml and contains 20% FeCl3 by mass. What is the molar concentration of FeCl3 in this solution?

Answer is 1.47 mol L^-1 (what does the -1 mean by the way?)

2. Which of the following is the major cause of hardness in natural groundwater?
A. Iron ions
*B. Calcium ions
C. Sodium ions
D. Chloride ions
E. Fluoride ions

3.Which of the following does not show any significant hydrogen bonding?
A. Ethyne, C2H2
B. Hydrazine, N2H4
C. Ethanol, C2H5OH
D. Hydrogen Fluoride, HF
E. Water, H2O

----------------------------

For #1, here's what I did: Molar mass of FeCl3 is 162, so since mass is 20% then 162*0.2=32.4g

n = mass / molar mass = 32.4/162 = 0.2

density = mass / volume
1.19 = 32.4 / volume
volume = 27.22 mL

Concentration = n / v = 0.2 / 27.22 = 0.007 M

What did I do wrong?

------

For #2, I narrowed it down to Sodium and Calcium from common sense, but I don't really know the chemistry behind it.

For #3, I'm lost. What does it mean by "significant" bonds - which concept is it testing?
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« Reply #5 on: May 07, 2009, 05:57:41 PM »

The density of the solution is 1.19g/ml which means that one litre would weigh 1190 g correct? Iron chloride makes up 20% of this mass, so you can find the mass of it. Then divide the mass by the molar mass and you'll get the number of moles, since there is already 1 litre, this is the concentration.
The mol L-1 just means moles per litre because you are dividing the the number of moles by the volume, i.e. moles/L. This will give mol L-1 just like if you did velocity=distance(m)/time(s) the unit for it will be ms-1.

For number two it is calcium ions.
http://en.wikipedia.org/wiki/Hard_water

Number three, the are only three elements that can form hydrogen bonds when bonded to hydrogen, they are nitrogen, oxygen and fluorine. Only one of the listed does not have these three.
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