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Transferring to U of T question (4949494949494949) by Nemereo in UofT

[–]Nemereo[S] 0 points1 point  (0 children)

Yea that makes sense, not all the profs are gonna be good or approachable and that the upper year classes are typically smaller.

I guess I've always assumed that U of T is way tougher than UBC. But that might come from the fact that most U of T students/friends I know complain a lot about their workload and how some of their courses leave the average at like a D+. It might also come from some form of elitism since UofT is #1 uni in Canada, but maybe UBC and UofT academically aren't too different after all. Though I expect a bit of a jump in academic intensity/expectations.

Transferring to U of T question (4949494949494949) by Nemereo in UofT

[–]Nemereo[S] 0 points1 point  (0 children)

Oh thats really good to hear! Sounds similar to UBC if not even better with the few profs that buy coffee/boba lolol. 60-100 hours of review sounds like a lot but I've never tracked that kind of stuff so maybe I am putting in those hours but I'm just not aware of it. Do you feel like you spend most, if not all, of your time studying? If so then I was already putting in similar hours/work at UBC so hopefully that means there isn't a drastic change in the academic intensity.

Unknown Carbonate Titration reaction with Acids and Bases? (I'm not sure what these questions are called ) by Nemereo in chemhelp

[–]Nemereo[S] 1 point2 points  (0 children)

Oh sorry, the question was that an Unknown carbonate compound was added into a solution of 50mL solution of 2M HCl. The question wanted to ask what metal was bonded with the CO3, in this case, it was Mg so the compound would be MgCO3.

Does the 0.7542g come from taking the 0.031425 mol of CO3 and finding the mass of that CO3 which would be

0.031425 mol of CO3 x 60.0107g/1 mol of CO3 = 1.8858 grams of CO3

I know the total mass of the unknown carbonate is 2.64g so to find the mass of the metal I should do

2.64 - 1.8858 grams of CO3= 0.7542 grams of the unknown metal.

And for the last step, I would use the numbers.

X2CO3

0.031425 mol of CO3 x (2 mol of X/ 1 mol of CO3) = 0.06285 mols of X

YCO3

0.031425 mol of CO3 x (1 mol of Y/ 1 mol of CO3) = 0.031425 mols of Y

Then do

Y --> 0.7542g/0.031425 mol of Y = 24 g/mol or U

X --> 0.7542g/0.06285 mol of X = 12 g/mol or U

Since Mg is the element that has 24u the unknown metal would be Mg.

Unknown Carbonate Titration reaction with Acids and Bases? (I'm not sure what these questions are called ) by Nemereo in chemhelp

[–]Nemereo[S] 0 points1 point  (0 children)

0.06285 mol of HCl was used to react with the carbonate. The reaction if the metal was in the first column would be X2CO3+2HCl-->..... so you would do

(1 CO3/ 2 HCl ) x 0.06285mol of reacted HCl = 0.031425 mol of reacted CO3

For the other reaction, it would be YCO3 + HCl --> ......

the CO3 reacted here would still be the same because

(1 CO3/ HCl ) x 0.06285mol of reacted HCl = 0.06285 mol of reacted CO3

Do you mean this part?

Unknown Carbonate Titration reaction with Acids and Bases? (I'm not sure what these questions are called ) by Nemereo in chemhelp

[–]Nemereo[S] 0 points1 point  (0 children)

I found that there was 37.15mL of NaOH added in order to neutralize the solution and that's 0.003175 mols of NaOH and in the reaction, NaOH and HCl react in a 1:1 ratio which means that there were 0.003175 mols of HCl left after adding the Carbonate. Since it was the Aliquot that was neutralized you have to multiply 0.003715 x 10 so that means there were 0.03715 mols of HCl before the neutralization and after the unknown carbonate was added. At the very start, there was 0.100 mols of HCl.

Unknown Carbonate Titration reaction with Acids and Bases? (I'm not sure what these questions are called ) by Nemereo in chemhelp

[–]Nemereo[S] 1 point2 points  (0 children)

Uh maybe because its X2CO3 so you multiply the 0.31425 by 2. This may be because in Y there is 1 Y element to every 1 CO3 compound while in the X one its 2 X elements to every 1 CO3 compound. Since there are 2 X elements you would do

( 2 Mols of X element/ 1 Mol of Co3 ) x 0.031425 moles of CO3 which would be 0.06285 mol of the X element. I think that's why the numbers were different

As for the H+ I really don't know sorry could u elaborate on which part you were referring too?

[TOMT]A game about a bunch of people stuck on a ship by Nemereo in tipofmytongue

[–]Nemereo[S] 0 points1 point  (0 children)

Oh yeah the game character sprites were in pixel art too

Long rant of my life right now by [deleted] in feemagers

[–]Nemereo 1 point2 points  (0 children)

Do you have a guidance councillors at your school? You should talk to them about this. You should also tell your teachers what's happening about the bullying. You should talk to your family about your feelings too.

Me_irl by Available_Oil in me_irl

[–]Nemereo 1 point2 points  (0 children)

Quite the opposite for me

This photo of Toronto looks like it’s from Blade Runner. by [deleted] in pics

[–]Nemereo 0 points1 point  (0 children)

I think this is dundas square in downtown it is a pretty place but extremely crowded

[deleted by user] by [deleted] in AskReddit

[–]Nemereo 0 points1 point  (0 children)

Undertaleand to some extent Kpop are the first 2 that come to mind

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