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Physics, Chemistry and Biology: Post your doubts here!

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In the rusting experiment, why does the water level rise in the test tube? I mean, does more water get formed or something? How? :/
 
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In the rusting experiment, why does the water level rise in the test tube? I mean, does more water get formed or something? How? :/
the oxygen in the test tube reacts with iron

this reduces the volume of oxygen gas making space for more water thus water level rises

if i m not correct then plz let me know
 
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so how do i know which one to say copper or cobalt
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?
 
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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. (y)
 
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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. (y)

i am sort of getting it ... so we assume that mass melted by heat from surroundings remain constant, right? That is how we can take the mass melted by heat from surroundings for the first 2 mins, same as for the second 2 mins ... correct?
 
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Guys please help.
M/J 2014 Physics Paper 62 Question 1 b and question 1 c(iii). :(

1.b) Rotate string around unsharpened section of pencil for a particular no. of times, say 3 turns. (We are taking more no. of turns for greater accuracy). Mark the string where 3 turns end and measure the distance. Now, divide that distance by 3 to get the circumference (which is length of one turn).

not sure about the other one ...
 
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This more challenging final part produced some good, we
Thank you so much! :) I'm stuck with the other one too. :/

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
 
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Thank you so much man. Are you giving the board exam for IGCSE Physics in M/J 2015? I have some doubts maybe we could help each other? :D Just sayin'
 
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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. (y)
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
Hi.. Can u please let me know that from where did u get access to the 2014 examiner's report..?! Please..!
 
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