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See, when you add 10mol/dm3 of NaOH to 10mol/dm3 of the acid...the concentration changes! because the volume change...so now NaOH has 10 mol/2 dm3...since we doubled the volume..thus 0.1/2 is 0.05 concntration!
Hmm, good question...!thanx bro....
but......look, when you're saying it's because of doubling the volume that the conc of NaOH changes, the ms is saying that 0.05 is the conc of the salt, not of the base (NaOH), (or is it somehow that the conc of the salt and the base is always the same, or something like that?)
that's the bit that i don't get, and the '0.075'.....i too still can't figure out how they get that
Perhaps i'm not making a lot of sense, but well, i'm so confused
Okay, but can you tell which year this question is in..oww okayy thanx, btw i have a test tomorrow morning, that's why i'm desperate :/ I hope i figure out the soln before then inshaAllah.
But thanx a million for your help
yes sure, it's on/11, paper 43.
I got it Alhamdulilah...!!
yea sorry, i meant acid...! Good luck with your test tomorrow, May Allah grant you success Ameen! Wa eyyakumwooow thanx bro! JazakAllah khair! I understood
PS: the part in red, you meant 0.05 from acid, nah?
Thank you......InshaAllah, Aameen
The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. Hydration enthalpies are always negative.
Hey you should read Chemistry AS Muhammad Ayub for this I Dont remember this but its written in that book so that u could understand
for q9: find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method then calculate the thermal energy absorbed by water wit Q=mc(delta thita)
hope it helped
q 4 :
Im still struggling a bit :/ i didn't get the "find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method", if you please give a little more explanation i will get the hang of it im sure.for q9: find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method then calculate the thermal energy absorbed by water wit Q=mc(delta thita)
sutbtract the amount of energy absorbed by the water from the total energy released and u'll get the thermal energy released in the expreiment. then from there calculate the amount of enrgy per gram of fuel burnt.
i'm sorry i don't have a calculator at hand so i just gave u the steps.... annyways u'll understtand better if u go through with the questions and work it out following the steps urself
And it is the right answerrrr. Thankyou thankyou veryy muchh i understoodq 4 :
calculate the total energy absorbed in bond breaking of CO and 2H2 :
1077 + 2(436) = 1949
calculate the total energy released in bond making of CH3OH:
3(410) + 360 + 460 = 2050
subtract the energy released from the energy absorbed in the reaction to calculate the enthaply change:
1949-2050 = -101kJ/mol
reply if u don't get anything and i'll xplain... and tell me if its the correct answer
umm... i don't exactly remember the formula or stuff
but i remember there was a way to do it with all the enthalpy changes like standard, combustion, hydration, etc. sab k different methods thay calculation k ... that's as far as i get this question and by the look of it i really don't wanna solve it if some1 gives the ryt answer and calculations plz tag me too ... lol
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