Can someone check this for me please?

[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