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More exothermic means releasing more energy when the reaction occurs. This would be a more negative value, like -3400 kJ/mol. A less exothermic reaction would release lesser heat in comparison to a more exothermic reaction. This would have a less negative value, e.g. -2000 kJ/mol is less exothermic than -3400kJ/mol.
thanks!
Are you sure that's an alkane and not an alkene?
Ethanal, It has hydrogen bonds present in itself because the last Oxygen is directly attached to a Hydrogen!
For the first query, the compound will be a carboxylic acid because it's heated under re-flux.Al SAlam Alaikum! please i have few questions in chemistry AS pleaseeeeee help me asap exam is tomorrowhttp://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s07_qp_2.pdf which is may/june 2007 paper 2 btw, question 2(d) please for the first column , what is this is compound and what will be its product on oxidation??Next is http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_qp_2.pdf which is may/june 2008 paper 2 , question 5 (a)ii and 5(e) what structural formula should we write can someone explain!!and please some one post Good chemistry AS notes so i can revise from it tomorrowOne more question was how can we find the bond angle when they ask us H-C-H for example and they say suggest bond angles
Thanks alotFor the first query, the compound will be a carboxylic acid because it's heated under re-flux.
For the second one, 5a(ii), it'll be oxidized to a carboxylic acid. For 5(e) it should have just -OH as the terminals. So that leaves us to fill up the bonds of Carbon. Hence C-C will be triple bonded, with OH at each of the terminals (as depicted by the marking scheme).
Go through your book, please! It has several chote chote important points that students tend to overlook!
You look for how many atoms are attached to the central one, and then how many lone pairs are there. Look in the Chp 3 or 4 closely in the Cambridge book, it has a table there!
(Y) exam in 1 hr and 40 mins from now :0 May Allah make it easy for all of us, All the best
Is the answer 2?
b) Each option gives a case choosing one of the three to be the slowest step, which the rate-determining step. The reactants of the equation in every case should appear in the rate equation if they affect the equilibrium. Taking step I, we have H2O2 + I– ---> IO– + H2O , this has H2O2 and iodide but no hydrogen ions. so the order for hydrogen would be zero, and that for the other two, 1.
No. 4 spots for solvent 1
haha ok dunno how to do that then
http://www.chemguide.co.uk/analysis/chromatography/paper.html this explains it really well.
b) Each option gives a case choosing one of the three to be the slowest step, which the rate-determining step. The reactants of the equation in every case should appear in the rate equation if they affect the equilibrium. Taking step I, we have H2O2 + I– ---> IO– + H2O , this has H2O2 and iodide but no hydrogen ions. so the order for hydrogen would be zero, and that for the other two, 1.
a=1, b=1, c=0
Step II: (IO-) + (H+) ---> HOI here there is hydrogen so for c =1, but the other reagent is an intermediate that *never* appear in a rate equation. the intermediate was formed using H2O2 and I- so the instead of IO-, these would appear i.e.
a=1 and b=1, c=1
Step III: The equation is HOI + H+ + I– ==> I2 + H2O
there is again an intermediate HOI, this was formed using H+ and IO-. This means that there would be two sources of hydrogen ions, one the third equation and the other the second equation, giving c=2. IO- takes us back to the first equation, which uses H2O2 and I-, leaving us again with two sources of I- ( the third eq. and the first one) so b=2. H2O2 comes only once, so a=1.
a=1, b=2, c=2.
Hope it helps.
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