Question

I have an unusual request, and I'm not sure if anyone would take it up, but I am in a terrible position and I don't know what to do. My chemistry teacher assigned an extremely difficult online hw assignment via a program called Aleks. I have worked 2 days straight without sleeping and I still can't get through everything. There were initially 100 questions (3 parts each), and I managed to finish all but 8, and I can only get credit if I finish everything. Would anyone be willing to do a screenshare or something and help guide me through the last questions. I urgently need help. I'll even pay.

Answer 1

No need to pay..just post your questions here. I'm sure if not me, someone else can help step by step that's what were all here for (: good luck!

Answer 2

oh man that would be fantastic. Thank you :). I think maybe the best way might be via screenshots. I've been so stressed because I'm on the verge of failing, and I've been trying as hard as I can to not retake this class :(. Thank you! I'll post some of the questions.

Answer 3

You can take screenshots of them and upload it as an attachment via "Attach File"

Answer 4

here's the one I've been stuck on... I have been struggling with acid-base questions. I just managed to figure out how to use ICE tables and K values... somewhat :/

Answer 5

what grade chem is this?

Answer 6

Alright give me like five minutes I'll see if I can solve it.

Answer 7

general chem (college level). It's just that the professor uses this online hw and it is terrible... you have to do every problem correctly 4 times in a row or you have to do it again...

Answer 8

thank you. I appreciate it... I will try again as well.

Answer 9

The sodium hydroxide (strong base) will neutralize the acetic acid (weak acid).\[NaOH + HC_2H_3O_2 \rightarrow NaC_2H_3O_2 + H_2O\]You are given the moles of NaOH. You are also given the volume & molarity of acetic acid, which you can use to find the moles of acetic acid; mol = molarity x volume. Use the stoichiometric ratios of the reaction to find the moles of acetic acid left and acetate formed.

The 2nd one is similar, except this time you have HCl neutralizing the weak base (KC2H3O2, which you can treat as C2H3O2 -, acetate). The original HC2H3O2 is just extra that you have to add on to the new HC2H3O2 formed.

Answer 10

Oh Bananas! I forgot about this :P Rougue got it (: