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the oxygen in the test tube reacts with ironIn the rusting experiment, why does the water level rise in the test tube? I mean, does more water get formed or something? How? :/
It also says DARK PINK liquid ( I think that's why as transition metals have coloured compounds)View attachment 51542
View attachment 51543
how it is COBALT nitrate
blue ppt was formed which is a test foe cu+2 ion
It also says DARK PINK liquid ( I think that's why as transition metals have coloured compounds)
It is a fact that most cobalt salts are pink just as most copper salts are blue and most nickel salts are green... so the fact that it forms a pink solution indicates it is cobalt.. btw which year's paper is this?so how do i know which one to say copper or cobalt
This might help u.. no need to learn allView attachment 51542
View attachment 51543
how it is COBALT nitrate
blue ppt was formed which is a test foe cu+2 ion
https://hgphysics.files.wordpress.com/2013/04/0625_w08_ma_31hg.pdf
page 9 .... question 5.c
Can someone plz explain why they are subtracting 2.1 g from 16.3 g?
TIA
Since they want to calculate latent heat required to heat the ice,
So when heater was switched off, the ice that melted due to energy FROM SURROUNDING was 2.1g.
Remember that when heated by the 40W heater, it's being heated by energy from heater as well as surrounding, and this is 16.3g.
So, to get an accurate result, we must use the mass that was heated by the heater ONLY.
So we subtract mass heated due to surrounding which is 2.1 from 16.3...
Hope it's clear enough, if you need any clarification, please ask.
Guys please help.
M/J 2014 Physics Paper 62 Question 1 b and question 1 c(iii).
Thank you so much! I'm stuck with the other one too. :/
Since they want to calculate latent heat required to heat the ice,
So when heater was switched off, the ice that melted due to energy FROM SURROUNDING was 2.1g.
Remember that when heated by the 40W heater, it's being heated by energy from heater as well as surrounding, and this is 16.3g.
So, to get an accurate result, we must use the mass that was heated by the heater ONLY.
So we subtract mass heated due to surrounding which is 2.1 from 16.3...
Hope it's clear enough, if you need any clarification, please ask.
Hi.. Can u please let me know that from where did u get access to the 2014 examiner's report..?! Please..!This more challenging final part produced some good, we
I think the following will help you. It's from the examiner report:
"ll-reasoned answers. The better candidates approximated the sharpened end to a cone and used an appropriate formula. Any sensible estimate was accepted, as long as candidates showed that the volume of the sharpened end was less than that of a cylinder of the same length. A number of candidates produced a volume estimate for the sharpened end which was larger than the volume of the unsharpened end, and made no comment that this could not possibly be so."
EDIT: Just to add on this, I tried it out and it does work! Use 1/3*pi*radius squared*height with values calculated earlier and volume is within range.
Cheers
Your image is printed on a smaller scale, maybe becoz of the marginsPlease some one help me with this problem. .. i have exam tomorrow
On the following question answer in mark scheme is 9.5.. but mine is coming exact 9
Can some one help me out with this? What can be the answer and why?View attachment 51656
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