Ask your own question, for FREE!
OpenStudy Feedback 16 Online
OpenStudy (anonymous):

when 50ml of gas of at 45 degrees celcius and 0.55atm is placed into 250ml flask the temperature is 32 degrees celcius, what is the new pressure?

OpenStudy (ghazi):

PV=nRT use it

OpenStudy (anonymous):

i dont know where to put what

OpenStudy (ghazi):

i assume number of moles of gas is constant

OpenStudy (ghazi):

\[ \frac{ P _{1} V _{1} }{ T _{1} }=\frac{ P _{2} V _{2}}{ T _{2} }\]

OpenStudy (anonymous):

there we go :D thanks! :)

OpenStudy (ghazi):

:D YW

OpenStudy (anonymous):

oh theres one more

OpenStudy (ghazi):

noooooooooooooo jk ;)

OpenStudy (anonymous):

When 0.60 grams of H2 are places into 2500.0 ml flask with an undisclosed amount of Bromine(+2), the total pressure is 12.5atm. if the temperature is 27 degrees Celsius, how many grams of bromine are in the flask?

OpenStudy (anonymous):

apparently im supposed to use more than just one gas law

OpenStudy (ghazi):

again i think you have to use PV=nRT

OpenStudy (ghazi):

you have, P,. you have to find n that is number of moles, you have temperature, you have gas constant that is R and you have volume just plug in and play :D

OpenStudy (anonymous):

yea but im supposed to use more than one law

OpenStudy (ghazi):

PV=nRT is just one law and earlier i made the comparison from one law in two cases

OpenStudy (anonymous):

okay can you explain what this formula is:\[[P_{ob}+a\left({n\over V}\right)^2]\times(v-nb)=nRT\]

OpenStudy (ghazi):

this is the general case of PV=nRT when your a and b=0 we get PV=nRT, hope you know what does a and b stands for

OpenStudy (anonymous):

lol, i dont

OpenStudy (ghazi):

awesome ;) google it :D lol

OpenStudy (ghazi):

n--> number of moles

OpenStudy (ghazi):

a and b are either determined empirically for each individual compound or estimated from the relations.

OpenStudy (anonymous):

i looked it up n now im even more confused..this is not helping :(

OpenStudy (ghazi):

\[a=\frac{ 27R^2T _{c}^2 }{ P _{c} }\]

OpenStudy (ghazi):

and \[b=\frac{ RT _{c} }{ 8P _{c} }\]

OpenStudy (ghazi):

Tc , Pc--> critical temperature and pressure

OpenStudy (anonymous):

question! the Volume in PV=nRT has t be in liters, right?

OpenStudy (ghazi):

umm yea

OpenStudy (anonymous):

u dont seem sure..?

OpenStudy (ghazi):

its been 5 years since i did it for the last time lol so kinda but generally you should use SI units so i think you can use liters or in this case go with ml , in physical chemistry we prefer ml

OpenStudy (anonymous):

okay well i figured that i need to find the pressure of H2 and subtract it from the total then use that left over pressure to calculate the moles of Br n convert that to grams to find grams...yea?

OpenStudy (ghazi):

yep :D if you have number of moles use \[n=\frac{ w }{ m }\] where w--> weight andn m--> molar mass

OpenStudy (anonymous):

okay well can you work it out with me? im not getting the right answer for some reason :S

OpenStudy (ghazi):

what is your number of moles?

OpenStudy (anonymous):

0.3 for H2

OpenStudy (ghazi):

then just plug in the values

OpenStudy (anonymous):

i did \[P_{H_2}(2500)=(0.3)(0.08206)(300)~~\implies~~P_{H_2}=0.00295416\]\[12.5-0.00295416=12.49704584\]\[(12.49704584)(2500)=n(0.08206)(300)~~\implies~~n_{Br}=1269.096377\]and i know that's not right, given the balance equation \(Br_2+2h_2\rightarrow2BrH_2\)

OpenStudy (anonymous):

@ghazi

OpenStudy (anonymous):

if you use liters though, it gives 0.97 moles which is 77.45 grams...i have no idea what to go by..??

OpenStudy (ghazi):

use liters

OpenStudy (ghazi):

whenever you have wrong answer try to check if your answer is reasonable or not

OpenStudy (anonymous):

thats what im trying to make sure of.. pressure of H2=2.954atm pressure of Br=9.56atm moles of H2=0.3mol moles of Br=0.97mol grams of H2=0.60g grams of Br=77.45g reasonable??

OpenStudy (ghazi):

yea this seems reasonable :)

OpenStudy (anonymous):

the question is worth 6pts out of 12...i cant afford it being wrong :P

OpenStudy (ghazi):

so i hope you got right :)

OpenStudy (anonymous):

k thanks a lot! :D

OpenStudy (ghazi):

:D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!