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Topic: International Olympiad preparation  (Read 18536 times)

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Online Borek

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Re: International Olympiad preparation
« Reply #30 on: May 03, 2013, 05:17:17 PM »
How many $1 coins do you need to pay if the price of the item is $2.7?
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Offline Big-Daddy

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Re: International Olympiad preparation
« Reply #31 on: May 03, 2013, 05:45:17 PM »
How many $1 coins do you need to pay if the price of the item is $2.7?

So then we have to round up :) 3 photons it is ... I'll get back to 2.3 at the next opportunity.

Offline Sophia7X

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Re: International Olympiad preparation
« Reply #32 on: May 03, 2013, 07:29:30 PM »
Raderford google 2013 IChO Scribd and you should find the problems.


« Last Edit: May 03, 2013, 07:39:40 PM by Sophia7X »
Entropy happens.

Offline Sophia7X

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Re: International Olympiad preparation
« Reply #33 on: May 03, 2013, 10:08:57 PM »
OK I got 27.3% for problem 2 #3 and 4(a) I got 3.47*10^7 m^3 for O2, and 4.26*10^7 kg not sure if correct.... anyone attempt these ones yet?
« Last Edit: May 03, 2013, 10:27:11 PM by Sophia7X »
Entropy happens.

Offline Rutherford

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Re: International Olympiad preparation
« Reply #34 on: May 04, 2013, 04:22:46 AM »
Thanks very much Sophia7X.
This problem becomes more and more strange. In 1. we need to calculate the minimum number of photons, but in 2. we find out that the Gibb's energy we used isn't correct.

If someone is interested in checking the answers for the first problem, I found a link few weeks ago: http://www.nanometer.ru/2013/02/06/13601765911974.html (it is in Russian >:D).

Offline Big-Daddy

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Re: International Olympiad preparation
« Reply #35 on: May 04, 2013, 07:50:55 AM »
Thanks very much Sophia7X.
This problem becomes more and more strange. In 1. we need to calculate the minimum number of photons, but in 2. we find out that the Gibb's energy we used isn't correct.

If someone is interested in checking the answers for the first problem, I found a link few weeks ago: http://www.nanometer.ru/2013/02/06/13601765911974.html (it is in Russian >:D).

Not quite ... we calculate standard Gibbs' energy ΔG° in 2.1, then used that for the minimum number of photons. In 2.2 though we calculate Gibbs' energy (not standard) ΔG, calculation of which relies on having ΔG° as I calculated a few posts ago. ΔGr° is the energy needing to be provided for the reaction to proceed, not ΔGr.
« Last Edit: May 04, 2013, 09:09:34 AM by Big-Daddy »

Offline Big-Daddy

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Re: International Olympiad preparation
« Reply #36 on: May 04, 2013, 07:51:55 AM »
OK I got 27.3% for problem 2 #3 and 4(a) I got 3.47*10^7 m^3 for O2, and 4.26*10^7 kg not sure if correct.... anyone attempt these ones yet?

Yeah 27.30% for 2.3, how did you arrive at those for 2.4?  We are not given how much solar energy is available ...

Offline Big-Daddy

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Re: International Olympiad preparation
« Reply #37 on: May 04, 2013, 10:44:09 AM »
OK I got 27.3% for problem 2 #3 and 4(a) I got 3.47*10^7 m^3 for O2, and 4.26*10^7 kg not sure if correct.... anyone attempt these ones yet?

OK so I just saw that actually we are given the solar energy available so we can calculate it. I don't think it's that vague.

2.4.a. N[photons]=8.53881·1033; m[CH2O]=4.2574·107 kg; V[O2]=3.46704·107 m3.
2.4.b. N[photons]=8.47896·1029; m[CH2O]=4,227.58 kg; V[O2]=3,442.7 m3.

Looks like it agrees completely with your calculations.

I can clarify any part of my method if someone wants.
« Last Edit: May 04, 2013, 12:03:32 PM by Big-Daddy »

Offline Big-Daddy

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Re: International Olympiad preparation
« Reply #38 on: May 04, 2013, 12:09:49 PM »
Am I misinterpreting 2.5 or is it unexpectedly easy?

E(absorbed by plants)=E(available)*(1/10)
E(converted)=E(absorbed by plants)*0.272966
E(converted)=E(available)*(1/10)*0.272966

All this from 2.4. And now the question asks for E(converted)/E(available).

E(converted)/E(available)=E(available)*(1/10)*0.272966/E(available)
E(converted)/E(available)=(1/10)*0.272966=2.73%

Which appears to have no dependence on E(available).

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