Can someone check this for me please?

  • Thread starter yoshi6
  • Start date
In summary: Ag^+ (aq) + 2e^- \rightarrow 2Ag (s) \ E^0 = (+0.80) * 22Ni^{2+} (aq) + 4e^- \rightarrow 2Ni (s) \ E^0 = (-0.28) * 2which then becomes2Ag^+ (aq) + 2e^- \rightarrow 2Ag (s) \ E^0 = +1.602Ni^{2+} (aq) + 4e^- \rightarrow 2Ni (s) \ E^0 = -0.56then we have to flip the anode
  • #36
yoshi6 said:
did i add them wrong?
yep

show me your work if you were to do it with the equation
 
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  • #37
Ecell = E cathode - E anode
= .80 - (-0.28)
=1.08 ?
 
  • #38
yoshi6 said:
Ecell = E cathode - E anode
= .80 - (-0.28)
=1.08 ?
correct

and if you just flip the equation and choose not to use the equation, you would've added and resulting in the same answer = +1.08
 
  • #39
okay, awesome...I think I have some studying to do before first year starts...
 
  • #40
yoshi6 said:
okay, awesome...I think I have some studying to do before first year starts...
alright, good luck! have fun
 
  • #41
yoshi6 said:
okay, awesome...I think I have some studying to do before first year starts...

chemistry
I also need to do some chemistry :cry:; I have been planning to do it since last two months.
What are you going into, and if you don't mind, where?

If you are planning to really do it; then, you may try
http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/organicx.htm

It has notes for all grade 12 topics/units!
 
  • #42
Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.
 
  • #43
yoshi6 said:
Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.
wow! beautiful campus ...
 
  • #44
you've been? Lol don't you live in texas?
 
  • #45
yoshi6 said:
you've been? Lol don't you live in texas?
i can be anywhere anytime ... <google> :biggrin:

it's nice tho, how's the weather? lol, it's way too hot down here.
 
  • #46
ah isee...it is a really nice campus...well, I seem to think it is really hot right now, it's about 31 celcius, what about over there?
 
  • #47
okay I have one last question that I am having trouble with. I am almost done my assignment. This is the last one...I have some examples in my book I am trying to follow, however, this is a little different.

If the Ka= 1.8 x 10^-5 for acetic acid (i.e . CH3COOH), what is the H+ ion concentration in a solution of this acid, if 1.2 grams of acid are dissolved in 1.0 L of solution.

So I assume I have to do this:

Ka= [H+][A-]/ [HA] ? and from there...
 
  • #48
so you have 1.2 grams of acetic acid, and you're told it's dissolved in 1.0L sol'n, what would your next step be?

your Ka expression is correct, question tho, is H+ and H3O+ the same?

by analyzing the Ka, what would you estimate the pH to be?
 
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  • #49
yes it is, isn't it?
 
  • #50
yoshi6 said:
yes it is, isn't it?
yes, what would you estimate your pH to be?

[tex]CH_3COOH (aq) + H_2O (l) \leftrightarrow H_3O^+ (aq) + CH_3COO^- (aq)[/tex]

what values you would input in your EQUIL for your ICE table?
 
  • #51
okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...
 
  • #52
okay that is supposed to be (9.0 x 10^-3)^2 divided by [CH3COOH]
 
  • #53
yoshi6 said:
okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...
where did you get 9.0 x 10^-3?
 
  • #54
wrong? I don't know I'm very confused, I am trying to follow the examples in my book, but I don't really understand what it going on...
 
  • #55
[tex]K_a=\frac{[H_3O^+][CH_3COO^-]}{CH_3COOH}[/tex]

[tex]1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]}{1.0}}[/tex]

you wrote your Ka correctly up there, idk what happened.

you're correct that the Acetate ion and Hydronium ion concentrations will be the same.
 
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  • #56
but where does the 60.06 come from? what is that
 
  • #57
yoshi6 said:
but where does the 60.06 come from? what is that
let me fix that, i should have that within another fraction

how do you calculate molarity? what units should/are used for equilibrium concentrations?
 
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  • #58
molarity = moles of solute/ liter of solution
 
  • #59
yoshi6 said:
molarity = moles of solute/ liter of solution
what is the molar mass of Acetic acid and the moles of 1.2 grams of Acetic acid that will be dissolved?
 
  • #60
60.05 g/mol i think
 
  • #61
yoshi6 said:
60.05 g/mol i think
correct, what about molarity and what would you do with it to find the Hydronium concentration at equilibrium?

[tex]K_a=\frac{[H_3O^+][CH_3COO^-]}{CH_3COOH}[/tex]

[tex]1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]}{1.0}}[/tex]

it could also be this

[tex]1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]-x}{1.0}}[/tex]
 
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  • #62
oh so that is where you got x^2/ [1.2/60.05]...
 
  • #63
yoshi6 said:
oh so that is where you got x^2/ [1.2/60.05]...
yep ... and have you read when you can neglect x = the concentration change?

i just calculated it and the difference was only by .09
 
  • #64
okay, makes sense, I'm just curious, what time is it over there?
 
  • #65
yoshi6 said:
okay, makes sense, I'm just curious, what time is it over there?
1AM, i'im so bored, lol. what time is it there?

i have class tomorrow but i can't sleep :-[
 
  • #66
wow...2am, I have been doing homework for hours...thank you so much for your help! I feel like I owe you a present
 
  • #67
yoshi6 said:
wow...2am, I have been doing homework for hours...thank you so much for your help! I feel like I owe you a present
naw idc, i want to be a chem tutor @ my school but i don't think i'll get hired b/c they prefer to hire professionals :( so I'm happy to help because i really don't want to forget what i learned, and i have no motivation to review on my own.

you should go to sleep, lol ...
 
  • #68
I should but I don't think I am going to yet...what school do you go to?
 
  • #69
yoshi6 said:
I should but I don't think I am going to yet...what school do you go to?
i go to a CC in Houston, lol ima PM u because i don't think they'd like chatting
 
  • #70
lol okay
 
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