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Lol so many answers lemme answer my question in short the one which has polar bonds shows the most deviation and is easily liquified and thanks evryone
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its CO2N2 because it has the biggest temporary induced dipoles out of all of them, due to having the most electrons. All the molecules there do not have an overall permanent dipole
oops my bad! didnt look at all the gassesits CO2
Please stop confusing just imagine what if this question appears in exam i will b like superconfused
lol if it dos come in da exam u will be luckiest prsn aliveits CO2
Please stop confusing just imagine what if this question appears in exam i will b like superconfused
btw i just made a mistake in dis so just wanted 2 post here incase any1 makes da same mistake
in da PV=nRT equation dont 4get 2 cnvrt cm3 by multiplying it wd (10^-6)...i 4got da cube part
anywayz BEST F LUCK evry1...plz do pray 4 an aswum ppr IA
for these questions u have 2 convert cm3 to m3 as m3 is da standard form.so cm to m is 10^-2 but its cm3 so to change it to m3 its (10^-6) ie into 3mmm do we always multiply cm^3 with 10^-6....lil confused here isn 1dm^3=1000cm^3 :O
use mcQthis may be a stupid question but can someone help me with this....Burning 1.00 g of octane produces 48.4 kJ of energy. Calculate the heat energy
produced by burning 1.00 g of methane ??
wooow thanks but i ws thinkin isn the volume taken as dm^3 as well ok fine its m^3for these questions u have 2 convert cm3 to m3 as m3 is da standard form.so cm to m is 10^-2 but its cm3 so to change it to m3 its (10^-6) ie into 3
yes ur ryt.volume is alwayz in dm3 but this is a GAS equation.here its m3, presure is in Pa and R is 8.314 and temperature is converted in k.if its in degree celcius just add 273 to change it 2 kelvinwooow thanks but i ws thinkin isn the volume taken as dm^3 as well ok fine its m^3
thank yuyes ur ryt.volume is alwayz in dm3 but this is a GAS equation.here its m3, presure is in Pa and R is 8.314 and temperature is converted in k.if its in degree celcius just add 273 to change it 2 kelvin
no problemthank yu
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