1. Some data is given below (i only extract the relevant parts)
Substance type of bonding mp/deg cel bp/deg cel
Iodine covalent 114 184
In the question, i was asked to name the substances that are NOT solids at r.t.p.
From the table, we know iodine IS a solid at r.t.p since it melts at 114deg cel. so i did not name iodine as not being a solid. But the answer did include iodine. Quite a controversial question since the iodine we use in the lab is a liquid.
2. May I know what are the conditions for esterification?
Is it H2SO4 as catalyst, 200 deg cel temperature and 1000 atm? (I might have mixed up the parts on temperature and pressure. Please correct.) As far as I know, the TB only tells us to use H2SO4. I did this question in my prelims but all the corrections are with my teacher so i can't check up the answer.However the answer in a similar question from TYS gives the conditions as contrn H2SO4 as catalyst and heat under reflux. For O level, which is the accepted answer?
3. Regarding the fractional distillation of crude oil, do we needa memorise the number of C atoms and boiling range of each fraction? (If I am not wrong, my teacher said there's no need to memorize. But I want to confirm this)
4.Here are 2 equations of the combustion of octane and hexdecane respectively
2C8H18 + 25O2 -> 16CO2 + 18H2O
2C16H34 + 49O2 -> 32CO2 + 34H2O
Using the equations, explain why hexdecane burns with a smokier flame than octane.
The model answer is: 1 mol of octane required only 12.5 mols of O2 for complete burning whereas 1 mol of hexadecane required 24.5 mol of O2 for complete combustion. Therefore, when hexadecane burns, more C is left unburnt making the flame smoky.
For me, I compared usin ghte no. of moles of CO2 produced. But after taking a look at the answer, I feel that my answer didn't really answer the question because indeed, if we have more C atoms left, it means that combustion is not as complete, hence more smoky reaction.
5. Solutions of 2 salts, A and B, were electrolysed using Carbon eletrodes. The following products were collected
Salt Products
A O2 and H2
B Cl2 and H2
Suggest the names of the 2 salts
I wrote the answer as aq. NaCl and concentrated NaCl for A and B respectively. But for the answer given, they list A as sodium sulfate. Are the namings strictly restricted to TWO DIFFERENT salts?
Thanks
1. please write the entire question. thx.
2. conc. H2SO4, rest correct
3. good to know
4. idk, for me, i also think it's CO2... but one possible explaination, incomplete combustion, CO(SMOKE can be seen)
5.i think it's not restricted.
Originally posted by anpanman:1. Some data is given below (i only extract the relevant parts)
Substance type of bonding mp/deg cel bp/deg cel
Iodine covalent 114 184
In the question, i was asked to name the substances that are NOT solids at r.t.p.
From the table, we know iodine IS a solid at r.t.p since it melts at 114deg cel. so i did not name iodine as not being a solid. But the answer did include iodine. Quite a controversial question since the iodine we use in the lab is a liquid.
2. May I know what are the conditions for esterification?
Is it H2SO4 as catalyst, 200 deg cel temperature and 1000 atm? (I might have mixed up the parts on temperature and pressure. Please correct.) As far as I know, the TB only tells us to use H2SO4. I did this question in my prelims but all the corrections are with my teacher so i can't check up the answer.However the answer in a similar question from TYS gives the conditions as contrn H2SO4 as catalyst and heat under reflux. For O level, which is the accepted answer?
3. Regarding the fractional distillation of crude oil, do we needa memorise the number of C atoms and boiling range of each fraction? (If I am not wrong, my teacher said there's no need to memorize. But I want to confirm this)
4.Here are 2 equations of the combustion of octane and hexdecane respectively
2C8H18 + 25O2 -> 16CO2 + 18H2O
2C16H34 + 49O2 -> 32CO2 + 34H2O
Using the equations, explain why hexdecane burns with a smokier flame than octane.
The model answer is: 1 mol of octane required only 12.5 mols of O2 for complete burning whereas 1 mol of hexadecane required 24.5 mol of O2 for complete combustion. Therefore, when hexadecane burns, more C is left unburnt making the flame smoky.
For me, I compared usin ghte no. of moles of CO2 produced. But after taking a look at the answer, I feel that my answer didn't really answer the question because indeed, if we have more C atoms left, it means that combustion is not as complete, hence more smoky reaction.
5. Solutions of 2 salts, A and B, were electrolysed using Carbon eletrodes. The following products were collected
Salt Products
A O2 and H2
B Cl2 and H2
Suggest the names of the 2 salts
I wrote the answer as aq. NaCl and concentrated NaCl for A and B respectively. But for the answer given, they list A as sodium sulfate. Are the namings strictly restricted to TWO DIFFERENT salts?
Thanks
1) The I2 can sublime at room temperature into I2 vapour
4) Unburnt C is what causes the smoke to be black. CO2 is colourless and thus does not affect the colour of the smoke regardless the amount.
no.3
if u looked at past year tys trends, u would notice that there is like almost no qn about that in the tys. so looking at what u ahve to study, do u feel that there is any point in studying it ?
of course la, as what transport said, it is 'good to know' but if it was me i will just learn the general trend.
no.2
actually u just folo tb is okay. because the tys ans written by other students also, and is not wrong. thou i will folo the tys de because it is done by uni student and is not wrong.
1) That liquid iodine in labs you use, it's called "Iodine Solution". Not "Iodine".
Originally posted by yiha093:no.2
actually u just folo tb is okay. because the tys ans written by other students also, and is not wrong. thou i will folo the tys de because it is done by uni student and is not wrong.
Actually, it's done by JC students. And a lot of the answers are wrong for the science TYS, though the Maths TYS are mainly accurate.
but the thing is right, if u r right, the ppl at cambridge cant fault u x,x
for question 2 pls note the word concentrated is very impt as without that it is not possible.
for question 3, just remember the fractions of crude oil (Peter Pan Never Kiss Dog Legs Before) can be a way to memorize it, you have to memorize the usage for the fractions. Eg naptha as the feedstock for chemical industry etc etc.
for question 4 if i aint wrong it should be calculate the percentage of carbon within the fuel, rather than the mole thingy. pls correct me if i am wrong in this,
for question 5, you do not expect the publisher to print down all the possible answers for you don't you? there is many many possible answers thats why
If iodine sublimes, how do we get the liquid iodine used in the lab? As for the smokier flame part, viewtyKU990 is correct, the more carbon, the smokier the flamer
Originally posted by dadeadman1337:If iodine sublimes, how do we get the liquid iodine used in the lab? As for the smokier flame part, viewtyKU990 is correct, the more carbon, the smokier the flamer
http://en.wikipedia.org/wiki/Lugol%27s_iodine
the lab iodine is iodine solution <---aqueous
not iodine liquid
iodine just put in water la . LOLLLL
you guys ask the funniest questions mannn
Q4. I won't tell you the answer. but I will guide you.
Compare C16H34 and C8H18. Which one contains more carbon, which one will require more energy to break?
Comparing C2H4 and C2H6, which one will require more energy to break? I would say it's C2H4, due to the presence of double bond. If you compare the alkene and alkane with 2 carbon atoms, alkenes always produces a smokier flame as compare to alkane.
hi, everyone, here's a question from 2006:
A liquid reacts with each of sodium carbonate, potassium hydroxide and ethanol. What is the liquid?
A) aqueous ammonia
B) ethanoic acid
C) ethyl ethanoate
D) hydrochloric acid
Ans: B
Why is it not hydrochloric acid? Thanks
because only dicarboxyic acid react with ethanol to form ester.
Originally posted by yiha093:because only dicarboxyic acid react with ethanol to form ester.
No. Not limited to di-carboxylic.
Originally posted by qdtimes2:hi, everyone, here's a question from 2006:
A liquid reacts with each of sodium carbonate, potassium hydroxide and ethanol. What is the liquid?
A) aqueous ammonia
B) ethanoic acid
C) ethyl ethanoate
D) hydrochloric acid
Ans: B
Why is it not hydrochloric acid? Thanks
Yes. Acids react will reaction with the above 3 substances. but for ethanol, it must be carboxylic acids (-COOH- group); so that it can form the -ate part of the ester. example ethyl ethanoate. This is in the O level syllabus.
At higher level, there are inorganic ester which are formed by reacting (phosphoric acid, nitric acid etc) with alcohol, for HCl I am not too sure about it.
I should have dropped to Combined Science when I had the chance ... Sigh.
Why does copper(II) oxide react with ammonia?
Originally posted by dadeadman1337:Why does copper(II) oxide react with ammonia?
see tb.
is that a reaction between an alkali and ammonium salt?
i checked wikipedia and it says that copper(II) oxide dissolve in aqueous ammonia. so it becomes an alkali is it ?
Originally posted by qdtimes2:is that a reaction between an alkali and ammonium salt?
i checked wikipedia and it says that copper(II) oxide dissolve in aqueous ammonia. so it becomes an alkali is it ?
alkali with ammonium salts gives ammonia guess, salt and water right? Think its in the textbook.
CuO dissolves in ammonia to form a salt. An alkali is a base dissolve in water by definition if never remember wrong and not an oxide in an alkali
copper (II) oxide dissolve in sodium hydroxide to form copper (II) hydroxide, hope thats help
Since it become copper (II) hydroxide, it is a base, not necessary alkali
Originally posted by qdtimes2:is that a reaction between an alkali and ammonium salt?
i checked wikipedia and it says that copper(II) oxide dissolve in aqueous ammonia. so it becomes an alkali is it ?
Aqueous Ammonia can react with CuO... recall about the test about Cation.
I observed that many of my students does not know the chemical formula for aqueous ammonia.
Anyone like to attempt or know the answer to which the formula of aqueous ammonia?
oh i see...thanks everyone
the formula of aqueous ammonia is NH4OH right?
but my teacher says that in the chemical equation where the question states only aqueous ammonia, we write only NH3 (aq).