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secondary 4 | A Maths
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LockB
LockB

secondary 4 chevron_right A Maths chevron_right Singapore

need help with part b, the correct answer is 1/8.dont really understand what they mean by "value of P changes by a factor of m"

Date Posted: 3 years ago
Views: 528
Eric Nicholas K
Eric Nicholas K
3 years ago
Part b is actually the same as part a if you look closely

When Q is actually halved from 1/3 to 1/6, the value of P actually changes from 2 to 1/4

To get from 2 to 1/4, we multiply 2 by 1/8

So m = 1/8
J
J
3 years ago
Not so evident in part (a) since the ⅛ does not actually appear, and it's only an evaluation of P by substituting values into the formula.

That's what part b) is there for. To test the concept.

Noticing that part (a) is the same is case of 'on hindsight' for most students.
Eric Nicholas K
Eric Nicholas K
3 years ago
I doubt it’s 1 mark in an O level paper otherwise, because similar questions are 2 marks.

I believe it’s a continuation of part i but they did not mention it explicitly.
J
J
3 years ago
@LockB what's the breakdown of the marks for this question?
Eric Nicholas K
Eric Nicholas K
3 years ago
It’s 2 marks for part a and 1 mark for part b.

Just nice I answered the exact same O level paper during a class yesterday afternoon.
LockB
LockB
3 years ago
yes it is 2m for part a and 1m for part b
LockB
LockB
3 years ago
by the way how to we know whether a substance is a reducing agent or oxidising agent? for example SO2 since i was not told to memorise it at all..

i am doing 2018 paper 2 now, it is in qn 1b. the question is "it changes the colour of potassium manganate(vii) from colourless to purple" but i cant do it as i dont know whether it is RA or OA, encountered this issue a few times already
Eric Nicholas K
Eric Nicholas K
3 years ago
I would expect SO2 to be a reducing agent which reacts with the oxidising agent KMnO4
J
J
3 years ago
I think you mean purple to colourless, LockB?

The manganese ion in potassium manganate (VII) (MnO4-) has an oxidation state of +7. The appearance is dark purple

This is reduced to Mn²+ (colourless appearance) when you have a reducing agent.

So SO2 is a reducing agent here and itself is oxidised to SO4²- ions sulfate/sulphate (most probably)
LockB
LockB
3 years ago
nope the paper stated colourless to purple... just realised it should be purple to colourless, maybe the question was set to trick us. or is colourless to purple also possible?
J
J
3 years ago
The purple to colourless observation for MnO4- ions being reduced to Mn²+ ions is something to be internalised.

You can easily work out that the other reactant is oxidised, which means that it is a reducing agent since it reduces the MnO4-
J
J
3 years ago
If colourless to purple, then it is the other way round.

Mn²+ ions are oxidised to MnO4- ions, so the SO2 is now an oxidising agent.
J
J
3 years ago
Anyway, I did a quick search.

https://m.facebook.com/keynotelearning/photos/pcb.2368545733174250/2368545529840937
This should answer your query.
Eric Nicholas K
Eric Nicholas K
3 years ago
It’s unusual for manganese to go from +2 or +4 to +7
J
J
3 years ago
I'll need to see the actual TYS question.

Could be an 'error' that the student was supposed to identify. Highly unlikely that Cambridge would make a mistake in the ordering of the words.
LockB
LockB
3 years ago
i will post the question at the answers
LockB
LockB
3 years ago
hi Mr Eric and Mr J! o level for emath, amath and chem has ended, thank you for your help in these subjects. it really helped me a lot in understanding these subjects and i have pretty much achieved constant As for these subjects. although i may not achieve A again for olevel, thank you for putting in much effort to clear my doubts, your explanations are better than my teacher's explanation :)
J
J
3 years ago
Welcome. See you around sometime in the future for higher level subjects
J
J
3 years ago
If you're taking Bio feel free to ask more questions, since Paper 1 is on tomorrow.

See 2 Answers

The word factor here basically means 'by how many times'
If I say P changes by a factor of 10, it means P is now 10 times as large as before.
So, if Q is multiplied by ½, and since P is proportional to Q³,
then P is multiplied by the cube of ½ i.e by (½)³ = ⅛
P is ⅛ as large as before. P changed by a factor of ⅛.
In your answer, you got it the other way round
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J
J's answer
1024 answers (A Helpful Person)
1st
LockB
LockB
3 years ago
thx :)
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LockB
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30 answers (A Helpful Person)
^ the question
LockB
LockB
3 years ago
part a is about identify the correct substance from a list of given substance hence it is not related
J
J
3 years ago
I see...

False then.

The colour change is from purple to colourless.

Potassium manganate (VII) can only be reduced, not oxidised further.

The oxidation state of Mn in the MnO4- is +7.

This is reduced to +2 (Mn²+ ions are formed)


SO2 is a reducing agent here.


This question part is to test whether you remember the colour change correctly or not.

Quite a basic question.
J
J
3 years ago
Anyway, the maximum oxidation state of Mn is +7.

You can think of it as Mn7+ ion, though usually these don't exist as standalone ions.

It typically exists in the form MnO4-
LockB
LockB
3 years ago
thx :) btw for the qn :
"iron from thr blast furnace contains carbon and silivon as impurities. what is the source of silicon impurities in the blast furnace?"

why must it be sand and not slag? since slag also contain silicon
J
J
3 years ago
Slag is a byproduct, not a source.

The origin of the silicon (Si) comes from the silicon dioxide SiO2 which makes up the majority of most forms of sand.