Chemistry: Post your doubts here!

[Metallic9896]

Bro, help..........

For question 23, there are ester bonds going on to an OH and COOH will be there (confirmed both by the COOC in the diagram and the fact that conc. sulfuric acid is being used). It can't be A for that reason, nor C. The debate now is between B and D. If you look at D, the left molecule, CH3COOH has only COOH, so it must in all cases react with the OH of the molecule on the right, HOCH2COOH. If that does happen, our product ends up looking as CH3COOCH2COOH, but this isn't a ring at all, so we can imagine that THIS molecule may react with HOCH2COOH again, but that ends up as CH3COOCH2COOCH2COOH, and this just shows that it keeps going on and on in a straight chain. On the other hand, if you look at B, we have just one TYPE of molecule but of course multiple of these in the whole material. So let's say two of these molecules react:

HOCH2COOH + HOCH2COOH

The red reacts with the red part and the blue with the blue. What do you end up with? You end up with a cyclic compound, that I can't draw here, but looks EXACTLY like the diagram. COOCH2COOCH2C It should look like this with the yellow C being the same (in a ring).

For question 21, you should recall instantly that one of the key differences between alcohols and carboxylic acids is that an alcohol reacts ONLY with a metal (sodium) to form a alcoxide ion (-1 charge), as it is not that strong of a oxidizing agent, whereas a carboxylic acid reacts like any acid with metals (sodium) alkalis (NaOH) metal carbonates and so on, to form the carboxylate ion (-1 charge). So when J is reacted with only sodium, the total -1 charges are 3, whereas with an alkali, it is -1. Since with an alkali it's only -1, we know that there's only ONE carboxylic acid group in J, automatically leaving us with TWO alcohol groups. From the 4 options, only C meets this condition. A has 2 carboxylic acid groups and 1 alochol, B has 1 each and an alkanal, D has no carboxylic acid, just an alcohol, ester, ether, and alkanal group.

For question 19, this bothered me too a bit when I first came across it. However, we should know that CO does NOT react spontaneously in air to form CO2, otherwise CO poisoning wouldn't really be a thing. B is something that I haven't heard of, same goes for D. C however is something we are expected to know from the Nitrogen and Sulphur chapter. NO2 does act as a catalyst (this actually came in my CIE finals paper 2 too) with SO2 and turns it into SO3 while itself beocming NO, which means that SO2 reduces NO2 to NO.

 

[Metallic9896]

few more questions...

36, catalytic converters are meant to turn NO and CO to N2 and CO2. You should know that. 1 is correct, quite obviously. 3 is definitely incorrect cause better conditions for combustion means greater chance of N2 forming NO or NO2 (or NO going to NO2) again, which kind of defeats the purpose. Similarly, 2 should be incorrect cause we want as much time as possible to be available for our substances to be in contact with the catalyst. Increasing the rate of flow would decrease the efficiency of the process.

12, I hate these kind of questions, but in our book, Cambridge International AS and A Level Chemistry Coursebook 2nd Edition, on page 166, it says that barium burns with a green color. So yeah, sometimes CIE like to give us completely rote-learning based questions like this too. You just have to know this.

 

[Metallic9896]

I hope I helped and I didn't tally my answer with the mark schemes so let me know if I'm wrong about anyone.

 

[Bishnu Dev]

I hope I helped and I didn't tally my answer with the mark schemes so let me know if I'm wrong about anyone.

It's correct! Thanks a lot bro!

 

[Bishnu Dev]

Help!

 

[Bishnu Dev]

Explain plz

 

[Bishnu Dev]

...................

 

[amina1300]

I have the components C A D for my As level should I give a retake ....?

 

[Bishnu Dev]

I have the components C A D for my As level should I give a retake ....?

Depends upon your targeted score

 

[Bishnu Dev]

Alright, Bishnu, typing my response to you.

Please explain the questions i have attached.......

 

[Metallic9896]

Please explain the questions i have attached.......

Again busy with college admissions, will get back to you ASAP this week or later.

 

[Bishnu Dev]

Again busy with college admissions, will get back to you ASAP this week or later.

ok bro

 

[Metallic9896]

ok bro

3. C6H12 when combusted completely would form 6 moles of CO2 and 6 moles of H2O. I'm sure you know this already. If not, then ask me. Anyway, through P the H2O is absorbed and through Q the CO2 is. Increase in masses P and Q are the masses of H2O and CO2 absorbed within them, respectively. Since mass = moles x Mr, mass of H2O that dissolved in P was 6 x 18 = 108, and mass of CO2 that dissolved in Q was 6 x 44 = 264. So mass of P increased by 108 g and mass of Q increased by 264 g. 108/264 = 0.41 or A.

40. Compound Q has an ester linkage in it. I hope you can see that (the COO-C part). When heating an ester with NaOH, you are doing an alkaline hydrolysis. The result should be an alochol at the -C part and an alkanoate salt at the COO- part. Since the right-most carbon is the alcohol side, the alcohol we'll get is CH3OH, while the rest of the compound will be the alkanoate as CH3CH2CH2COO-Na+, these are 1 and 2. 3 cannot be formed, so answer should be B.

12. The graph at higher temp. should move slightly towards the right and have a more flattened peak. Answer is B from mark scheme, but this is kinda shady, would be appreciated if any other member could help us out on this one.

18. NH3 forming NH4+ is clearly using the lone pair on N so B is out. A is out cause the same is happening just that instead of NH4+ we are forming CH3NH3+, CH3+ is receiving the lone pair of N from NH3 rather than H+. D is out cause it's a transition metal complex, and although you won't know it in AS too well, but in A2 we know for a fact that if NH3 is involved in such a complex it is donating it's lone pair to form that complex, it's acting as a ligand. This leaves us with C, here a salt is formed, which is neutral. Since Na is Na+ for certain, NH2 is NH2- ion. Since the lone, non-banding pair of N gets involved when going from NH3 to NH4, I'm sure when going from NH3 to NH2, that's not the case, we are in fact losing an H+, which then forms an IONIC bond with Na+.

BTW for Questions 12 and 18 you're literally venturing into 13 year old past papers, I think it's best to stick and keep revising the latest 5-6 years one first. I got an A* in Chem and never had to go back that far, just a little advice. If you've already done them then by all means do these lol, but at the end, come back to the latest ones.

34. Very intersting question. I think what they'rre trying to get at is make you analyze the entire situation and see which options direct relate to the issue. So first let's talk about what's going on. They tell us that Hydroxyapatite, Ca5(PO4)3OH is more or less tooth enamel. That's fact 1.

Fact 2 is the set of equations given in saliva. In the first equation tooth enamel breaks to form Ca+, PO43-, and OH-. In the second equation reading from right to left, PO43- reacts with H+ to form HPO42-.

Now our concern is understanding why high presence of H+ will PROMOTE dissolving of tooth enamel. In other words, we want to explain why H+ will make the first equation go to the forward direction (as that direction involves breaking down of tooth enamel).

First statement says that OH- react with H+ to neutralize. This SUPPORTS our concern because when we add lots of H+, the OH- will react with it, thus, the equilibrium of the first equation will shift towards the forward direction. In other words, H+ will react with OH-, decreasing the conc. of OH-, thus making the equilibrium shift towards the right side to make more OH- to compensate for the ones tha were lost by reaction with H+, hence dissolving the tooth enamel.

Second statement says that PO43- accepts H+, this is fromt he second equation. What does this mean for tooth ename? This also SUPPORTS dissolving of tooth enamel. How? Well this is how: When we add lots of H+, as per second equation, there will be an increase in conc. of H+, making the equilibrium shift towards left, making PO43- react with the extra H+ to form HPO42-. Now when PO43- will react, what does this mean for first equation? This means that conc. of PO43- is decreasing, implying that in the first equation, the conc. of products is decreasing, encouraging the first equation equilibrium to shift towards the forward direction, and thus, like statement 1, making tooth enamel dissolve, to create more PO43-, which will in turn react with H+ to form HPO42-. Make sense?

Third statement says that Ca+ will react with acids (HPO42- in this case). This is going against dissolving of enamel because the Ca+ will in fact react with HPO42- and directly conflict with statement 2's reaction, as it will encourage equilibrium of second equation to want to go back to the right side.

This is how I see it. Anyone else feel free to correct me and explain further if I'm wrong.

7. The first row of the table is a strong acid strong base reaction so clearly the most heat would be evolved by such a reaction as maximum neutralization is going to take place, the heat being 57 kJ/mol. The fourth row ALSO has a 57 kJ/mol enthalpy change, and the acid is nitric acid, a strong acid, hence, the base must be a strong base too. This eliminates options B and D because R is assumed as ammonia there, which is a weak base. Now we have to choose between A and C. Q is the same in both and this gives us a clue that, as per our assumption, a strong acid and weak base (the third row, hydrochloric acid a strong acid, while btoh A and C say that Q, the base of that row, is ammonia, a weak base) should give an enthalpy change lower than 57 kJ/mol. Based on this, since the second row ALSO has an enthalpy change LESS than 57 kJ/mol, one of the two must be "weak", since the base is already given as sodium hydroxide, a strong base, then P, the acid MUST be weak. This is only in A where the acid suggested is ethanoic acid. It cannot be C because it suggests sulphuric acid, and it is a strong acid, so if that was the case, the enthalpy change shuold have been 57 kJ/mol, but it's not, thus proving A to be the correct option.

I hope I helped.

 

[Bishnu Dev]

Third statement says that Ca+ will react with acids (HPO42- in this case). This is going against dissolving of enamel because the Ca+ will in fact react with HPO42- and directly conflict with statement 2's reaction, as it will encourage equilibrium of second equation to want to go back to the right side.

Thanks bro!
I didn't get this part. Why can't Ca++ react with acid? and How does it conflict the second statement? If Ca2+ is reacting with HPO42-, it will encourage the reaction to shift to Left in equation 2, isn't it?

 

[Metallic9896]

Thanks bro!
I didn't get this part. Why can't Ca++ react with acid? and How does it conflict the second statement? If Ca2+ is reacting with HPO42-, it will encourage the reaction to shift to Left in equation 2, isn't it?

Oh, my bad. It's not that. I wasn't thinking clearly while writing that. The reason is, HPO42- is NOT an acid. It's hydrogen phosphate. You can google this too. It's the conjugate base of phosphoric acid and is basic in nature. https://www.quora.com/Why-is-HPO4-hydrogen-phosphate-not-phosphoric-acid Hence, the third statement simply does not apply.

 

[Lee Qian Yi]

Calculate the pH of the buffer formed when 10 cm3 of 0.1 m0ldm-3 NaOH is added to 10 cm3 of 0.25 moldm-3 CH3COOH, whose pKa=4.76
ANS : pH=4.58
How to calculate ? I know that need to use this formula pH= pKa + log ( conc. salt / conc. acid ) , but how to find the conc. of both solvents ?

 

[Metanoia]

Calculate the pH of the buffer formed when 10 cm3 of 0.1 m0ldm-3 NaOH is added to 10 cm3 of 0.25 moldm-3 CH3COOH, whose pKa=4.76
ANS : pH=4.58
How to calculate ? I know that need to use this formula pH= pKa + log ( conc. salt / conc. acid ) , but how to find the conc. of both solvents ?

Use a table to keep track of the moles of CH3COOH removed by the NaOH , and thus the amount of salt formed.

You should be able to find the respective concentrations easily after that

 

[krishnapatelzz]

 

[Holmes]

I have the components C A D for my As level should I give a retake ....?

What is your overall grade of Chemistry?
If it is B then ............ don't
Because you can improve in A2 as well: where P4 has 38.5% weightage ,you know
So take your decision calmly and wisely.
hope it helps a little.

 

[Bishnu Dev]

help