Chemistry: Post your doubts here!

[HELLNO]

yes you should write it/label it as divided flask It's actually a standard laboratory equipment.

Yeah I was just kidding
Best of luck!!!!

 

[unutbenit]

thank you

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

 

[msucurt]

Someone please help if you dont mind. I know its something simple, just having difficulty seeing what Ive done wrong. Based off the marking scheme, Im not sure why the number of moles HCl is .008.

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.

• 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)

thanks!!

 

[Shaaaazxx]

Can someone please help me with these questions?

 

[Shaaaazxx]

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 !

i dont understand this...

 

[Cadence]

The

Can someone please help me with these questions?

answer for the first one is potassium?

 

[Shaaaazxx]

The

answer for the first one is potassium?

nope, it's sodium

 

[Bah...]

Hello,
Can somebody please help me with the structure? I'm a bit confused.



This is what the ms says



Thank you

 

[areeba240]

can some plzz explain?
9701_w16_qp_41

 

[darks]

Help me understand this Please!!
ms
anode, O2,O2,Br2
cathode, Ag,H2, H2
Thanks!

 

[HELLNO]

View attachment 62129
Help me understand this Please!!
ms
anode, O2,O2,Br2
cathode, Ag,H2, H2
Thanks!

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

 

[fahadhameedahmad]

higher potential gains electrons at cathode

lower potential loses electrons at anode

www.youtube.com/fahadsacademyonline
www.fahadsacademy.com

 

[Metanoia]

i dont understand this...

Perhaps you can list down the other option(s) you thought were also correct, then we can guide you accordingly.

We can try replacing option A (correct answer) with something familiar

CH4 (g) --> C (g) + 4 H (g) ∆H

Bond energy of C-H is ∆H/4
1/4 CH4 (g) --> 1/4C (g) + H (g)

H4 (g) --> C (g) + 4 H (g) ∆H

Analogously
XYn (g) --> X (g) + n Y (g) ∆H

Bond energy of X-Y is ∆H/n
1/n XYn (g) --> 1/n X (g) + Y (g) ∆H

 

[Metanoia]

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.

• 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)

thanks!!

Check the information under "PART A) METHOD"

FB3 (diluted HCl) is prepared from diluting FB2 (concentrated HCl)

concentration of FB3 = (10 x 2)/250 = 0.08 mol/dm^3

 

[Metanoia]

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

A thistle funnel can be used

 

[darks]

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

Thanks A lot!

 

[darks]

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.

 

[HELLNO]

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.

(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.

Ag+ + e- ==> Ag has an E*cell of 0.8 v the backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode (E*cell=positive electrode-negative electrode)

The Ag+ concentration decreases in the solution, favouring the backward reaction (Le Chatelier's principle) The backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode



(iii) In Fe3+(aq) + Fe2+(aq) no change will take place, because its just gonna increase the sulphate ions in the solution which has zero effect.

While in Ag2SO4(aq) it will decrease the Ecell, thats because the sulphate ions precipitate the the Ag+ forming Ag2SO4, therefore it decreases the Ag+ concentration favouring the backward reaction, as explained above ^^

(iv) Basically what I said in (iii)

 

[areeba240]

9701_w09_qp_42
qtn 5(a)
can someone plzz explain me this part?

 

[areeba240]

can some plzz explain?
9701_w16_qp_41