Can someone check this for me please?

[rocomath]

did i add them wrong?

yep

show me your work if you were to do it with the equation

 

[yoshi6]

Ecell = E cathode - E anode
= .80 - (-0.28)
=1.08 ?

 

[rocomath]

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

 

[yoshi6]

okay, awesome...I think I have some studying to do before first year starts...

 

[rocomath]

okay, awesome...I think I have some studying to do before first year starts...

alright, good luck! have fun

 

[rootX]

okay, awesome...I think I have some studying to do before first year starts...

chemistry :yuck:
I also need to do some chemistry ; 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!

 

[yoshi6]

Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.

 

[rocomath]

Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.

wow! beautiful campus ...

 

[yoshi6]

you've been? Lol don't you live in texas?

 

[rocomath]

you've been? Lol don't you live in texas?

i can be anywhere anytime ... <google>

it's nice tho, how's the weather? lol, it's way too hot down here.

 

[yoshi6]

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?

 

[yoshi6]

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...

 

[rocomath]

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?

 

[yoshi6]

yes it is, isn't it?

 

[rocomath]

yes it is, isn't it?

yes, what would you estimate your pH to be?

$$CH_3COOH (aq) + H_2O (l) \leftrightarrow H_3O^+ (aq) + CH_3COO^- (aq)$$

what values you would input in your EQUIL for your ICE table?

 

[yoshi6]

okay

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

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

I think...

 

[yoshi6]

okay that is supposed to be (9.0 x 10^-3)^2 divided by [CH3COOH]

 

[rocomath]

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?

 

[yoshi6]

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...

 

[rocomath]

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

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

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.

 

[yoshi6]

but where does the 60.06 come from? what is that

 

[rocomath]

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?

 

[yoshi6]

molarity = moles of solute/ liter of solution

 

[rocomath]

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?

 

[yoshi6]

60.05 g/mol i think

 

[rocomath]

60.05 g/mol i think

correct, what about molarity and what would you do with it to find the Hydronium concentration at equilibrium?

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

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

it could also be this

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

 

[yoshi6]

oh so that is where you got x^2/ [1.2/60.05]...

 

[rocomath]

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

 

[yoshi6]

okay, makes sense, I'm just curious, what time is it over there?

 

[rocomath]

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 :-[

 

[yoshi6]

wow...2am, I have been doing homework for hours...thank you so much for your help! I feel like I owe you a present

 

[rocomath]

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 ...

 

[yoshi6]

I should but I don't think I am going to yet...what school do you go to?

 

[rocomath]

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

 

[yoshi6]

lol okay