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secondary 4 | A Maths
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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"
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
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.
I believe it’s a continuation of part i but they did not mention it explicitly.
Just nice I answered the exact same O level paper during a class yesterday afternoon.
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
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)
You can easily work out that the other reactant is oxidised, which means that it is a reducing agent since it reduces the MnO4-
Mn²+ ions are oxidised to MnO4- ions, so the SO2 is now an oxidising agent.
https://m.facebook.com/keynotelearning/photos/pcb.2368545733174250/2368545529840937
This should answer your query.
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.
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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
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.
You can think of it as Mn7+ ion, though usually these don't exist as standalone ions.
It typically exists in the form MnO4-
"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
The origin of the silicon (Si) comes from the silicon dioxide SiO2 which makes up the majority of most forms of sand.