Chemistry
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OpenStudy (anonymous):
At 1 atm , if in the mercury coloum , mercury was replaced by water the height of the coloum would be?
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mathslover (mathslover):
what do u want to calculate @Yahoo!
mathslover (mathslover):
and may i know that what you tried?
OpenStudy (anonymous):
we need to find the height of mercuty colum
mathslover (mathslover):
ok so what did u try ?
OpenStudy (anonymous):
i think there is some formula
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OpenStudy (anonymous):
but i cant remember that
OpenStudy (anonymous):
@mathslover
mathslover (mathslover):
wait
OpenStudy (anonymous):
pressure = density x g x height
OpenStudy (anonymous):
density of water is 1.00g/cm3
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mathslover (mathslover):
pressure is also given 1atm ... g is 9.8 m\s^2
mathslover (mathslover):
solve for h now
OpenStudy (anonymous):
@mathslover but the asnwer should be 10336mm
mathslover (mathslover):
1 meter = how much mili meter ?
OpenStudy (anonymous):
1000
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mathslover (mathslover):
1.00g/cm3 = 1g / 1000 mm^3
OpenStudy (anonymous):
@mathslover r u stuck?
mathslover (mathslover):
yes too much :(
mathslover (mathslover):
h1d1g = h2d2g
why are we not comparing that of water to mercury ?
OpenStudy (maheshmeghwal9):
sorry dude me no chem:/
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mathslover (mathslover):
so u have got the answer
OpenStudy (anonymous):
@mathslover we dont have h1 and h2 sooo
mathslover (mathslover):
ok ...
OpenStudy (anonymous):
Nop in that they have given the height
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OpenStudy (anonymous):
i linked to get idea for u
mathslover (mathslover):
wait i am searching for a better idea
OpenStudy (anonymous):
ok
mathslover (mathslover):
1 atmosphere = 760 mmHg
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
@mathslover u got it
mathslover (mathslover):
yes ... some what
OpenStudy (anonymous):
yup can u show that!!!
mathslover (mathslover):
that is all written in comments lolz :( sorry i am trying a lot
OpenStudy (anonymous):
wat u mean?
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mathslover (mathslover):
@quarkine is there .. no worries now
OpenStudy (anonymous):
P2-P1= d*g*h
|dw:1339935408324:dw|
so h can be calculated as we know d(density),g and Pa=P2...so what is the question??
@mathslover :will try..