Chemistry: Post your doubts here!

[syed babar]

Hi guys,
As there is less than a month left in the result, the anxiety is off the roof. Can any of you guys predict the threshold for A* in chemistry 9701/42 & 52 taken this year and 12, 22, 33 taken last year. I had an A in AS, and p5 went amazingg almost 27 or above. P4 even was okayish. Expecting a 75 in p4. Will that be enough for an A*?

 

[Metallic9896]

Yeah more than enough bud don't worry about it.

 

[Dukula Jayasinghe]

Hi people,
I have prepared a lattice enthalpy examination paper (A2 only). If you would like to get it marked, do the paper and scan or take photos and send it back to me personally through [email protected]

 

[Wei Ling]

Can anyone explain to me why doesn't HCL react with KI to produce chlorine gas?

 

[Bishnu Dev]

Can anyone explain to me why doesn't HCL react with KI to produce chlorine gas?

CL is more electronegative than I so Iodine in KI cannot replace CL from HCL.....

 

[Bishnu Dev]

Please explain..................

 

[Bishnu Dev]

No problem! Feel free to ask more via pm or on here.

 

[Deeksha Tamang]

why when copper reacted with clorine only cucl2 is formed?

 

[Metallic9896]

Please explain..................

For Q32, we know that the first statement is true simply because of the existence of ONA at the end as well as the entire sulphate group. It's used as a detergent too so it makes sense some of it must be soluble, or so I think. 2 is correct because alkyl chains are hydrophobic, so while they don't dissolve in water, they do dissolve in organic solvents like oil droplets. Finally, 3 is correct because as you can see all carbons are attached to 4 other atoms so it must be tetrahedral. All 3 statements are correct. Answer is A.

For Q33 you have to picture the diagram for Hess' law. When you atomize graphite or diamond, you will end up with the same product i.e. gaseous carbon atoms. So, imagine that to get to gaseous carbon atoms, you have two routes, one from graphite directly to gaseous carbon atoms, the other from graphite to diamond and then to gaseous carbon atoms:

Graphite --- > Carbon atoms
Graphite --- > Diamond --- > Carbon atoms

Since graphite to diamond is endothermic, and since atomisation itself is also always endothermic, this means that when going from graphite to diamond, some of the energy needed to go from graphite to carbon is already added, and then the rest is used to go from diamond to carbon. In other words, according to Hess' law, if you start with graphite and end with carbon atoms, then the total energy must be the same, let's call it X. Let's call diamond to carbon atoms Y, and graphite to diamond is given as 3.

Graphite -- > Carbon atoms = X kJ/mol

Graphite --- > Diamond = +3 kJ/mol --- > Carbon atoms = Y kJ/mol

X kJ/mol = +3 + Y kJ/mol

Thus, X kJ/mol is greater than Y kJ/mol, i.e. graphite's direct atomisation is greater than diamond's atomisation. This makes the first statement correct.

For the second statement, again, it's another version of atomisation. Since we need to add MORE energy to break bonds in graphite compared to diamond, this means that the bond enthalpy of graphite IS greater than the bond enthalpy of diamond.

For the third statement, we know that combusion is always exothermic, and you have to imagine another Hess' law diagram in your mind. The end product for the combustion of either graphite or diamond is CO2, and the enthalpy change is negative. Thus,

Graphite --- > CO2 = -A kJ/mol
Graphite --- > Diamond = +3 kJ/mol --- > CO2 = -B kJ/mol

-A kJ/mol = +3 - B kJ/mol

Thus, since the MAGNITUDE of the CHANGE is being referred, you can multiply the equation by -1 to consider the magnitudes. This turns it into:

A kJ/mol = -3 + B kJ/mol

SInce you have to subtract from B to get A, this means B as a CHANGE of enthalpy in MAGNITUDE is greater than A, making the third statement correct too. If they asked which one of the two is MORE negative or LESSER in VALUE, then it would ALSO be B, but that means that the CHANGE in enthalpy of B is GREATER than A. Don't let the signs confuse you and make sure you understand this perfectly. All 3 statements are correct. Answer is A.

For question 14, I don't think you need to thinkt too hard about it. Overthinking can cause problems too. In reaction 1 all we're doing is DECOMPOSING a compound, so that means you ADDED heat to BREAK it down, which means it's endothermic. In reaction 2, in all honesty this is where I say that the book recommended by CIE is HIGHLY HIGHLY valuable. I wish I knew about it in my A Level, I wouldn't have bothered going to class. Reaction 2 is described WORD for WORD in the Cambridge book, it says:

Calcium oxide, CaO, reacts with water to form calcium hydroxide. If water is dripped onto the surface of a lump of calcium oxide it causes a vigorous reaction. It gives off so much heat that some of the water boils off as the solid lump appears to expand and cracks open.

As you can see, it gives off heat and is exothermic. Even the reaction 1 is described as:

The carbonates and nitrates of the Group 2 elements decompose when heated.

Thus, reaction 1 is endothermic, 2 is exothermic. And the quotations are taken from pages 166 and 168 of Cambridge International AS and A Level Chemistry Coursebook Second Edition by Lawrie Ryan and Roger Norris. Get this book, I can gaurantee you no question will come from outside of it. As much as we A Level kids tend to assume that FSc kids only get book questions, and although the majority of A Level questions ARE conceptual, but that little difference that'll get you the A* is only made up by learning from the book directly too.

 

[Bishnu Dev]

For Q32, we know that the first statement is true simply because of the existence of ONA at the end as well as the entire sulphate group. It's used as a detergent too so it makes sense some of it must be soluble, or so I think. 2 is correct because alkyl chains are hydrophobic, so while they don't dissolve in water, they do dissolve in organic solvents like oil droplets. Finally, 3 is correct because as you can see all carbons are attached to 4 other atoms so it must be tetrahedral. All 3 statements are correct. Answer is A.

For Q33 you have to picture the diagram for Hess' law. When you atomize graphite or diamond, you will end up with the same product i.e. gaseous carbon atoms. So, imagine that to get to gaseous carbon atoms, you have two routes, one from graphite directly to gaseous carbon atoms, the other from graphite to diamond and then to gaseous carbon atoms:

Graphite --- > Carbon atoms
Graphite --- > Diamond --- > Carbon atoms

Since graphite to diamond is endothermic, and since atomisation itself is also always endothermic, this means that when going from graphite to diamond, some of the energy needed to go from graphite to carbon is already added, and then the rest is used to go from diamond to carbon. In other words, according to Hess' law, if you start with graphite and end with carbon atoms, then the total energy must be the same, let's call it X. Let's call diamond to carbon atoms Y, and graphite to diamond is given as 3.

Graphite -- > Carbon atoms = X kJ/mol

Graphite --- > Diamond = +3 kJ/mol --- > Carbon atoms = Y kJ/mol

X kJ/mol = +3 + Y kJ/mol

Thus, X kJ/mol is greater than Y kJ/mol, i.e. graphite's direct atomisation is greater than diamond's atomisation. This makes the first statement correct.

For the second statement, again, it's another version of atomisation. Since we need to add MORE energy to break bonds in graphite compared to diamond, this means that the bond enthalpy of graphite IS greater than the bond enthalpy of diamond.

For the third statement, we know that combusion is always exothermic, and you have to imagine another Hess' law diagram in your mind. The end product for the combustion of either graphite or diamond is CO2, and the enthalpy change is negative. Thus,

Graphite --- > CO2 = -A kJ/mol
Graphite --- > Diamond = +3 kJ/mol --- > CO2 = -B kJ/mol

-A kJ/mol = +3 - B kJ/mol

Thus, since the MAGNITUDE of the CHANGE is being referred, you can multiply the equation by -1 to consider the magnitudes. This turns it into:

A kJ/mol = -3 + B kJ/mol

SInce you have to subtract from B to get A, this means B as a CHANGE of enthalpy in MAGNITUDE is greater than A, making the third statement correct too. If they asked which one of the two is MORE negative or LESSER in VALUE, then it would ALSO be B, but that means that the CHANGE in enthalpy of B is GREATER than A. Don't let the signs confuse you and make sure you understand this perfectly. All 3 statements are correct. Answer is A.

For question 14, I don't think you need to thinkt too hard about it. Overthinking can cause problems too. In reaction 1 all we're doing is DECOMPOSING a compound, so that means you ADDED heat to BREAK it down, which means it's endothermic. In reaction 2, in all honesty this is where I say that the book recommended by CIE is HIGHLY HIGHLY valuable. I wish I knew about it in my A Level, I wouldn't have bothered going to class. Reaction 2 is described WORD for WORD in the Cambridge book, it says:

Calcium oxide, CaO, reacts with water to form calcium hydroxide. If water is dripped onto the surface of a lump of calcium oxide it causes a vigorous reaction. It gives off so much heat that some of the water boils off as the solid lump appears to expand and cracks open.

As you can see, it gives off heat and is exothermic. Even the reaction 1 is described as:

The carbonates and nitrates of the Group 2 elements decompose when heated.

Thus, reaction 1 is endothermic, 2 is exothermic. And the quotations are taken from pages 166 and 168 of Cambridge International AS and A Level Chemistry Coursebook Second Edition by Lawrie Ryan and Roger Norris. Get this book, I can gaurantee you no question will come from outside of it. As much as we A Level kids tend to assume that FSc kids only get book questions, and although the majority of A Level questions ARE conceptual, but that little difference that'll get you the A* is only made up by learning from the book directly too.

Thanks bro

 

[Bishnu Dev]

Please explain these two

 

[Bishnu Dev]

For Q32, we know that the first statement is true simply because of the existence of ONA at the end as well as the entire sulphate group. It's used as a detergent too so it makes sense some of it must be soluble, or so I think. 2 is correct because alkyl chains are hydrophobic, so while they don't dissolve in water, they do dissolve in organic solvents like oil droplets. Finally, 3 is correct because as you can see all carbons are attached to 4 other atoms so it must be tetrahedral. All 3 statements are correct. Answer is A.

For Q33 you have to picture the diagram for Hess' law. When you atomize graphite or diamond, you will end up with the same product i.e. gaseous carbon atoms. So, imagine that to get to gaseous carbon atoms, you have two routes, one from graphite directly to gaseous carbon atoms, the other from graphite to diamond and then to gaseous carbon atoms:

Graphite --- > Carbon atoms
Graphite --- > Diamond --- > Carbon atoms

Since graphite to diamond is endothermic, and since atomisation itself is also always endothermic, this means that when going from graphite to diamond, some of the energy needed to go from graphite to carbon is already added, and then the rest is used to go from diamond to carbon. In other words, according to Hess' law, if you start with graphite and end with carbon atoms, then the total energy must be the same, let's call it X. Let's call diamond to carbon atoms Y, and graphite to diamond is given as 3.

Graphite -- > Carbon atoms = X kJ/mol

Graphite --- > Diamond = +3 kJ/mol --- > Carbon atoms = Y kJ/mol

X kJ/mol = +3 + Y kJ/mol

Thus, X kJ/mol is greater than Y kJ/mol, i.e. graphite's direct atomisation is greater than diamond's atomisation. This makes the first statement correct.

For the second statement, again, it's another version of atomisation. Since we need to add MORE energy to break bonds in graphite compared to diamond, this means that the bond enthalpy of graphite IS greater than the bond enthalpy of diamond.

For the third statement, we know that combusion is always exothermic, and you have to imagine another Hess' law diagram in your mind. The end product for the combustion of either graphite or diamond is CO2, and the enthalpy change is negative. Thus,

Graphite --- > CO2 = -A kJ/mol
Graphite --- > Diamond = +3 kJ/mol --- > CO2 = -B kJ/mol

-A kJ/mol = +3 - B kJ/mol

Thus, since the MAGNITUDE of the CHANGE is being referred, you can multiply the equation by -1 to consider the magnitudes. This turns it into:

A kJ/mol = -3 + B kJ/mol

SInce you have to subtract from B to get A, this means B as a CHANGE of enthalpy in MAGNITUDE is greater than A, making the third statement correct too. If they asked which one of the two is MORE negative or LESSER in VALUE, then it would ALSO be B, but that means that the CHANGE in enthalpy of B is GREATER than A. Don't let the signs confuse you and make sure you understand this perfectly. All 3 statements are correct. Answer is A.

For question 14, I don't think you need to thinkt too hard about it. Overthinking can cause problems too. In reaction 1 all we're doing is DECOMPOSING a compound, so that means you ADDED heat to BREAK it down, which means it's endothermic. In reaction 2, in all honesty this is where I say that the book recommended by CIE is HIGHLY HIGHLY valuable. I wish I knew about it in my A Level, I wouldn't have bothered going to class. Reaction 2 is described WORD for WORD in the Cambridge book, it says:

Calcium oxide, CaO, reacts with water to form calcium hydroxide. If water is dripped onto the surface of a lump of calcium oxide it causes a vigorous reaction. It gives off so much heat that some of the water boils off as the solid lump appears to expand and cracks open.

As you can see, it gives off heat and is exothermic. Even the reaction 1 is described as:

The carbonates and nitrates of the Group 2 elements decompose when heated.

Thus, reaction 1 is endothermic, 2 is exothermic. And the quotations are taken from pages 166 and 168 of Cambridge International AS and A Level Chemistry Coursebook Second Edition by Lawrie Ryan and Roger Norris. Get this book, I can gaurantee you no question will come from outside of it. As much as we A Level kids tend to assume that FSc kids only get book questions, and although the majority of A Level questions ARE conceptual, but that little difference that'll get you the A* is only made up by learning from the book directly too.

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

 

[SalmaM1]

Help!!

 

[Metallic9896]

I will get back to both of you soon.

 

[Bishnu Dev]

few more questions...

 

[Metallic9896]

Help!! View attachment 62629

First you since you asked only one question. For questions like this it's best to do some working as you read the question so that you have everything laid out in front of you. We know there's a white powder than contains both magnesium oxide and aluminium oxide. We also know that 100/1000 x 2 = 0.2 moles of NaOH causes aluminium oxide in X grams of the mass to dissolve. From the first equation given we know that for every 2 moles of OH-, 1 mole of Al2O3 dissolves, so since we used 0.2 moles of NaOH, 0.1 moles of Al2O3 dissolved. Hence, Aluminium oxide present in x grams of the mixture = 0.1 moles of aluminium oxide.

800/1000 x 2 = 1.6 moles of HCl causes the entire mass of the white powder to dissolve, which means BOTH the 0.1 moles of Al2O3 AND the unknown moles of MgO. Here the situation isn't THAT straightforward. We know the TOTAL moles of HCl that reacted, but we don't know HOW much HCl reacted with EACH of the two oxides. We DO however know, that Al2O3 is 0.1 moles. Using that information, AND the second equation given, we can first figure out how much HCl reacted with the Al2O3, subtract it from 1.6 to find out the remaining moles of HCl, use the third equation to find out the moles of MgO that is present in x grams of the white powder. So let's do that now. From the second equation we see that 6 moles of H+ react with 1 mole of Al2O3, and we also know that 1 mole of HCl contains 1 mole of H+, so that's something that makes it a bit easier (as was the case above when 1 mole of NaOH contained 1 mole of OH-). We know that the aluminium oxide is 0.1 moles, so from the equation

Al2O3 : H+ / HCl
1 : 6
0.1 : y

Find y, and it comes out to be 0.6 moles of H+. This means that from the 1.6 moles that we had, 0.6 moles of HCl reacted with Al2O3, leaving behind 1 mole of HCl to react with MgO. Using the third equation we can then find how many moles of MgO were present. From the equation, 2 moles of HCl react with 1 mole of MgO, so 1 mole of HCl must react with 0.5 moles of MgO. Hence, the correct answer is D, 0.10 moles of Al2O3 and 0.50 moles of MgO. I hope this helped.

 

[Metallic9896]

Alright, Bishnu, typing my response to you.

 

[Metallic9896]

Please explain these two

For the first question, it's from 2011, and if you see the syllabus (http://theallpapers.com/papers/CIE/AS_and_ALevel/Chemistry (9701)/9701_y11_sy.pdf) we see that in Nitrogen and Sulphur, the section of sulphur contains the following points:

(g) explain why atmospheric oxides of nitrogen are pollutants, including their catalytic role in the oxidation of atmospheric sulfur dioxide
(h) describe the formation of atmospheric sulfur dioxide from the combustion of sulfur contaminated carbonaceous fuels
(i) state the role of sulfur dioxide in the formation of acid rain and describe the main environmental consequences of acid rain
(j) state the main details of the Contact process for sulfuric acid production
(k) understand the industrial importance of sulfuric acid
(l) describe the use of sulfur dioxide in food preservation


On the other hand, our current syllabus only has 2 points:

a) describe the formation of atmospheric sulfur dioxide from the
combustion of sulfur-contaminated fossil fuels
b) state the role of sulfur dioxide in the formation of acid rain and describe
the main environmental consequences of acid rain

Clearly, we don't need to know as much about sulphur as 2011 candidates needed to. However, I'll still try to answer this. We know it isn't B cause with it it's only as an acid that leaves behind a salt of Magnesium Sulphate. It isn't A either cause although not explicit in the syllabus I've noticed that CIE do expect us to know that ethanol is dehydrated to ethene, so like no oxidation going on there.

With propanenitrile we know that sulphuric acid turn nitriles to carboxylic acids, so we have HYDROLYSIS going on there. This leaves us with D and even if we don't get why D, we did eliminate other options. But if you think about D, it's NaBr. Even in our current syllabus, in the topic of Group VII, we read

describe and explain the reactions of halide ions with:
(ii) concentrated sulfuric acid

With the bromide ion in NaBr, we know that the bromide ion is continously oxidized from Br- to Br2 as . How it reacts with NaBr as an acid is that H2SO4 reacts with NaBr to form HBr and HNaSO4, which means it donates a proton, so that's that. Everythig is explained here http://www.chemguide.co.uk/inorganic/group7/halideions.html

For the second question just see that the last subshell to be filled is the p subshell and since it's COMPLETELY filled, remember that it is a nobel gas, so answer is D. You don't have to bother about anything else.

 

[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.