Cathode_RT
I'm so done with trimesters bruh
seems like a lot of us finished today on chem lol,
feels like nobody here does physics???
seems like a lot of us finished today on chem lol,
feels like nobody here does physics???
yuh yuh like 7.8 or smthn was the number and the magnitude was really smallYeah I got something to the -11 but it looked really small in comparison to the other given values so don't know if i did it right
ill be like the 5th person to say this but yes i got 81.7 L alsoyeah same I got 81.7 but was very skeptical because it seemed way too high
I've still got physics to goYay thanks.
Also carrots I'm gonna miss seeing u you're like everywhere Was chem your last exam?
I think I got a mass of ammonium that was in between 0.1 and 1 but i dont remember the exact mass.Also for the question where they found the nh4 in the fertiliser i had no clue what to do. i had 1.55g or smth. I didnt know what to do at the step when they boiled the ammonia with NaOH
I think I got that but like 10^-11 maybei got like 7.3 X10^-9 it was 7. smth
I’ve noticed ketones and aldehydes being treated as functional group isomers in trial papers asw so im not really sure, it’s definitely odd since they’re both carbonyl but maybe they’re treated as functional group isomers in the syllabus or something?PV = nRT is the way to go for Q27 (CO2 production from fermentation) because the gas volumes on the data sheet only apply to temperatures at 273 K and 298 K. Alternatively, if you used one of those (since the pressure matched) you would then need to adjust the volume using a Charles' Law calculation.
Question 21 is badly written as there is (technically) no correct answer. Butanone and butanal are NOT functional group isomers as they have the same functional group - carbonyl. They are different classes of compounds (a ketone and an aldehyde), but that is not the same thing as classifying them as having different functional groups. In the same way, 1-butanol, 2-butanol, and 2-methyl-2-propanol all have the same functional group (a hydroxyl) but have some differences in chemistry due to the different positioning of the functional group. In this case, though, we call all of them alcohols and consider them part of the same class.
I doubt it will cause a problem - what they wanted was obvious - but it is technically wrong.
Well, the syllabus and the Department of Education / NESA / etc do things their own way... and trials often have questionable content... but functional groups and compound classes are not the same thing.I’ve noticed ketones and aldehydes being treated as functional group isomers in trial papers asw so im not really sure, it’s definitely odd since they’re both carbonyl but maybe they’re treated as functional group isomers in the syllabus or something?
An alkene Q will go to dichloroalkane RFor compound R how did you know where to put the two chlorines? And what was the answer
The addition of chlorine to an alkene produces a dichloroalkane.For compound R how did you know where to put the two chlorines? And what was the answer
I mean like how do we know it's not 1,1 or 2,2 or 1,3 ?The addition of chlorine to an alkene produces a dichloroalkane.
The fact that the mass spectrum appears to show only one is misleading. The product should be 1,2-dichloropropane.
The mass spectrum of the compound can be seen at this link: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78875&Mask=200#Mass-Spec
doesnt 1,2-dichloroprop-1-ene fit cause you could assume Q might have a double bond and since doesnt specify excess cl2 it could only have 1 double bond being saturated also 1,2-dichloroprop-1-ene has 32 percent carbon by mass and has molecular ion peak at 114 it also has a fragment at 15 and 62?The addition of chlorine to an alkene produces a dichloroalkane.
The fact that the mass spectrum appears to show only one is misleading. The product should be 1,2-dichloropropane.
The mass spectrum of the compound can be seen at this link: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78875&Mask=200#Mass-Spec
You can see that it matches the spectrum provided in the exam, but with the lowest intensity peaks missing.
Taking the 35Cl : 37Cl ratio as 3 : 1 yields the parent (molecular) ion, [C3H6Cl2]+ . at m/z = 112, 114, and 116 in a 9 : 6 : 1 ratio.
Rather than wait for some genius person who got everything right, how bout we just compile everyone’s answers.Solutions ?
Surely multiple choice at least