Şərhlər
28 iyul, 2015
SarvarHello, when the Olympiad tasks be published?
29 iyul, 2015SCWithin a week
Name
Write a reply
26 iyul, 2015
DaveDear Science Commitee
When will the exams published?
Thank you
Dave
When will the exams published?
Thank you
Dave
31 iyul, 2015DaveHow about the final score of the olympiad?
26 iyul, 2015SCImediately after the Olympiad
Name
Write a reply
25 iyul, 2015
imranWhen result will be announced?
26 iyul, 2015SCAt the Closing ceremony
Name
Write a reply
18 iyul, 2015
FarrukhPlease look for problem 12 part 1 answerwhy there are 5 isomers. If you look at 3d structure cis-diammine-cis-dichloro-cis-dibromo-platinum(4) has optical isomer
21 iyul, 2015SCYes, certainly. Only geometric isomers were taken into consideration.
Name
Write a reply
17 iyul, 2015
Alexander BaumgartnerWill it be necessary to memorise the spectrochemical and trans-effect series, or will enough information be provided on the exams?
20 iyul, 2015SCIt is better to memorise the main members of both series. More detailed information will be given if necessary.
Name
Write a reply
14 iyul, 2015
itaiIn the solution to problem 11 you draw the d-orbitals splitting of Fe3+ as high spin (but Fe2+ as LS), though CN is strong field ligand, and higher oxidations tend to be at LS. Therefore, can you please explain why did you draw Fe3+ as HS? thanks
15 iyul, 2015SCThe Fe3+ is really HS in Prussian blue but not Fe2+ because of the coordination mode of ligand: the former is coordinated by N, the latter - by C. The CN group gives high splitting (LS-complexes) when it is C-coordinated.
Name
Write a reply
13 iyul, 2015
Alexander BaumgartnerHello
On problem 23, question 2, the solution mentions that k = l+m. This is assuming that the quantity of hydrogen is equal to that of the sum of the two halogens.
Should the solution not mention the possibility that the quantitiy of hydrogen atoms could be less than the quantity of halogens? Chlorine gas is moderately soluble in water, so couldn't that have formed instead of HCl?
On problem 23, question 2, the solution mentions that k = l+m. This is assuming that the quantity of hydrogen is equal to that of the sum of the two halogens.
Should the solution not mention the possibility that the quantitiy of hydrogen atoms could be less than the quantity of halogens? Chlorine gas is moderately soluble in water, so couldn't that have formed instead of HCl?
14 iyul, 2015SCIt's problem 26. Chlorine does not fit the condition that gas is colorless.
Name
Write a reply
13 iyul, 2015
AMIn problem 24, there is 11 fold decrease volume due to absorption of which gas? Is it due to CO2 or N2 or both?
17 iyul, 2015XWhy did you divide by N2
14 iyul, 2015SC(CO2+N2) : N2 = 11
Name
Write a reply
12 iyul, 2015
ICould you please elaborate on the meaning/significance of the number density in second part of problem 6. Please and thank you.
13 iyul, 2015SCThe number density in this case is the total number of particles adsorbed on the glass slide divided by the area. It is equal to the probability that a given area 1 sq.micrometer contains an adsorbed particle.
Name
Write a reply
12 iyul, 2015
JanetWhat's the model of the calculator? Could I use my personal one or the provided calculators are mandatory?
13 iyul, 2015TlThe calculator on the picture does not look exactly like Casio fx-85MS. It looks like Casio fx-991MS
12 iyul, 2015SCCasio fx-85MS. Personal calculators are not allowed, the organizers provide all the participants with the official calculator
Name
Write a reply
12 iyul, 2015
DomIn problem 3 part 4 how is the number of atoms greater than 3. The answer is really vague. Could you please elaborate?
12 iyul, 2015SCCv = 3R is possible only for non-linear molecule. The exact number of atoms cannot be determined.
Name
Write a reply
10 iyul, 2015
Abdullah HemburkerHello! Regarding problem 13, questions 2 and 3 why is the gold being oxidesided to the +1 oxidation state even though we have proven in the first question that this state is unstable and disproportionates into elemental gold and gold(III)?
11 iyul, 2015SCIn these questions, oxidtaion state of gold is not specified. In any case, oxidation starts with Au --> Au(+1)
Name
Write a reply
8 iyul, 2015
JakeIn part 1.1 of problem 5, why is the change in number of moles of glucose not equal to 1/6 C6H12O6?
9 iyul, 2015SCThe change of chemical variable is the change of moles divided by stoichiometric coefficient, which for glucose is 1/6
Name
Write a reply
6 iyul, 2015
JanetIs it known what eddition of Excel would be presented in the olympiad? They differ from one another somewhat, so it would be easier to focus on only one version while practicing.
9 iyul, 2015SCThe versions of the software we propose are Windows 8 and Excel 2013.
Name
Write a reply
6 iyul, 2015
jWhy on problem 12 part 3 when you have a group with high trans effect like an alkene it is substituted? Thanks.
9 iyul, 2015SCThe alkene is substituted by thiourea as we use an excess of the ligand and reflux the mixture making the volatile alkene to evolve.
Name
Write a reply
1 iyul, 2015
kHi! Problem 26 can you explain why s is equal to [Br]?
2 iyul, 2015SCBecause all dissolved AgBr dissociates into Ag+ and Br-
Name
Write a reply
1 iyul, 2015
FelixProblem 5.5 c robs as a function of DeltaGreaction.
The solutions propose some function with an slope going to infinity by converging of DeltaGreaction --> 0
But this is not possible for the e-function? it has to be an slower increasing function.
My second question is Is it possible to continue the drawing of this function for postiv DeltaGreaction values. I think it is.
Despite we would get negativ robs-values
The solutions propose some function with an slope going to infinity by converging of DeltaGreaction --> 0
But this is not possible for the e-function? it has to be an slower increasing function.
My second question is Is it possible to continue the drawing of this function for postiv DeltaGreaction values. I think it is.
Despite we would get negativ robs-values
2 iyul, 2015SC1) No infinte slope - neither in the figure, nor in the solution. The formula is: r_obs = r1*(1-exp(DeltaG/RT)). The slope at zero is -r1/RT.2) Negative robs means that the reverse reaction prevails - this is true because DelatG > 0.
Name
Write a reply
1 iyul, 2015
KychytkyCan you explain please, why E^0'(2BrO3/Br2)=E^0(2Bro3/Br2)-0.059*12/10*pH problem 14.2
2 iyul, 2015SCApply the Nernst equation to the half-reaction 2BrO3- + 12H+ + 10e --> Br2 + 6H2O
Name
Write a reply
30 iyun, 2015
AIn Problem 33, what are the concentrations of MnO4 -1?
2 iyul, 2015SCVariable concentrations, about 10^(-4) M - see the table at p. 67 of Prep Problems
Name
Write a reply
28 iyun, 2015
DanielOn problem 16, 3.1 the a part, in which you have to calculate the titre of the solution, it is said that the quantity of water a molecule of iodine can titrate is in a relationship 11 shouldn't it be 21 ?
2 iyul, 2015SCThe correct relation is 1:1. H2SO4 in a reaction between I2 and SO2 is formed only in water solutions when H2O is in excess. See https://en.wikipedia.org/wiki/Karl_Fischer_titration
1 iyul, 2015Daniel... relationship 1 to 1 shouldn't it be 2 to 1 ? *
Name
Write a reply