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Yeah I was just kiddingyes you should write it/label it as divided flask It's actually a standard laboratory equipment.
Best of luck!!!!
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Yeah I was just kiddingyes you should write it/label it as divided flask It's actually a standard laboratory equipment.
Moles of Iodine in the original solution= 0.2*50/1000=1x10-2
Moles of Iodine in the 10cm3 portion=1x10-2*(10/100)=1x10-3
The ratio of iodine:thiosulfate is 1:2
Therefore number of moles of thiosulfate= 1x10-3*2=2x10-3
Volume of 0.1 mol/dm3 thiosulfate= (2x10-3/0.1)*1000 (From the formula, volume=(moles*1000)/concentration)
Answer is 20 cm3
Could someone please help with a calculation from Paper 3 (may/june_2009). Its the calculations portion of question #1. Ive attached my work. The question is a titration of borax vs HCl.
thanks!!
- I am having difficulty understanding where the .008moles HCl come from on the mark scheme on #1c(i).
- I am having difficulty understanding why the Mr is so large from my calculations on #1c(iv)
Remember what bond energy is,it is the energy released when a covalent bond of a gaseous molecule is broken to give gaseous atoms and not molecules or compounds or molecules. So in A only is this happening in the rest of the options ∆H not only contains the bond breaking energy but also bond formation so the resulting ∆H for B,C and D is lesser than the value of X----Y bond energy !
I hope you get it its pretty simple just try to understand and ask any question no matter how childish it is about this explanation !
answer for the first one is potassium?Can someone please help me with these questions?
The
answer for the first one is potassium?
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be subject to electrolysis. Thus, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. In this case we have two possible reductions and two possible oxidations, listed here along with their reduction or oxidation potentials:View attachment 62129
Help me understand this Please!!
ms
anode, O2,O2,Br2
cathode, Ag,H2, H2
Thanks!
i dont understand this...
Could someone please help with a calculation from Paper 3 (may/june_2009). Its the calculations portion of question #1. Ive attached my work. The question is a titration of borax vs HCl.
thanks!!
- I am having difficulty understanding where the .008moles HCl come from on the mark scheme on #1c(i).
- I am having difficulty understanding why the Mr is so large from my calculations on #1c(iv)
Though 2 years aren't much but I did like from 2010, and the lastest ones are more important, so you'd do fine.
Did you do all variants?
Also any idea, how do we keep reactants separate until the reaction starts?
like maybe tie a thread to crucible and put the bung. then use thread to empty it. I think I read something like that
Thanks A lot!AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be subject to electrolysis. Thus, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. In this case we have two possible reductions and two possible oxidations, listed here along with their reduction or oxidation potentials:
Reduction:
2H+ + 2e- ==> H2, E = 0.00 V
Ag+ + e- ==> Ag, E = 0.8 V
Oxidation:
2F- ==> F2 + 2e-, E = -2.87 V
2H2O ==> O2 + 4H+ + 4e-, E = -1.23 V
(ALL THE VALUES ARE FROM THE DATA BOOKLET)
The reactions with the *most positive* potentials are most likely to proceed. Therefore, the reduction of Ag+, which occurs at the cathode, is the most favorable reduction reaction and the oxidation of H2O, which occurs at the anode, is the most favorable oxidation reaction. Consequently, the major products of this hydrolysis are Ag and O2, whereas H2 and F2 are not produced in appreciable amounts.
Same principle with all three
(ii) At standard conditions the concentration should be 1mol/dm3, in this case the Ag+ concentration as given in the beginning of the question is 2.5×10–2 mol/dm3, which is lower than 1.View attachment 62130
Please explain!! thanks. All parts after E*cell I know it's a lot but this comes a lot in exam and my concepts of this topic are not so good.
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