Chemistry: Post your doubts here!

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Metallic9896

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Bishnu Dev said:
ok bro :)
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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

Bishnu Dev

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Metallic9896 said:
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.
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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?
 
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Metallic9896

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Bishnu Dev said:
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?
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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.
 
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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

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Lee Qian Yi said:
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 ?
Click to expand...

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

pH.PNG
 
Holmes

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amina1300 said:
I have the components C A D for my As level should I give a retake ....?
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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. :)
 
i_try9621

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Capture.JPG I need help with this. From March 2017 Paper 4 I only get 6 different carbon environments
 
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