Question

i need someone to look over my chem answers please

Answer 1

5) calculate the percent of acetic acid in the vingar( the density of vingar is 1.002g/mL)
*im not really sure how to do this

Answer 2

Yeah, I don't know how to do this. Sorry.

Answer 3

Oh gosh, this looks hard. I haven't took the type of chemistry with moles, and equations etc.

However, I could balance equations, if you need help with that?

Sorry. :(

Answer 4

Yeah @Photon336

Answer 5

He's an expert ^

Answer 6

1. = CH3COOH + NaHCO3 = CH3COO-Na+ + H2O + CO2
2. = Correct.
3-5 I dont know
6. = Correct.
7. = HBr is a strong acid, ionizes completely;

Mole = (0.250 mol/L )( 0.0264 L) = 0.0066 mol

HBr> H+ + Br-
0.250 M . 0.250 M 0.250 M
0.0066 mol .. 0.0066 mol


H+ + OH- > H2O
0.0066 mol H+ neutralizes 0.0066 mol OH^- ion

CsOH is a strong base and it dissociates completely;

CsOH(aq) > Cs^+(aq) + OH^-(aq)
0.0066 mol < 0.0066 mol

M = mol / Volume = 0.0066 mol / 0.0300 L = 0.22 M

Answer 7

1st-
the oxygen are not balanced
check again :)

Answer 8

2nd=correct

Answer 9

i agree with obamacare
and i think we need some more info for questions 3-5

Answer 10

yeah

Answer 11

did they say that the molarity of given vinegar is 1.61M?

Answer 12

it just says calculate the molarity of the vineger sample

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @song_of_the_sole
4) calculate the number of grams of CH3COOH in the vinegar
* 1000mL=1L=\(\color{red}{1.61}\) 1.61 mol acetic acid =48g acetic acid
9.86mL=9.86/1000*1.61=0.158746=.76g acetic acid
\(\color{blue}{\text{End of Quote}}\)
then why did you take molarity as 1.61M? :~)

Answer 14

well
if your teacher asked you to take molarity as 1.61M then its alright

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @song_of_the_sole
3) calculate the molarity of the vinegar sample.(dont forget to convert mL to L)
*1 mole
\(\color{blue}{\text{End of Quote}}\)
but here you said that the molarity of vinegar sample is 1

and then you take it as 1.61

Answer 16

i mean like
why did you say that molarity is 1 even when your told you that its 1.61

Answer 17

yes
\(\color{blue}{\text{Originally Posted by}}\) @song_of_the_sole
3) calculate the molarity of the vinegar sample.(dont forget to convert mL to L)
*1 mole
\(\color{blue}{\text{End of Quote}}\)
*\(\color{red}{1.61M}\)

and then if molarity is 1.61M and volume is 9.86mL then moles will be 0.01 moles
so

\(\color{blue}{\text{Originally Posted by}}\) @song_of_the_sole
2)calculate the number of moles of NaHCO3 that were required to neutralize the CH3COOH in the vingar.
*1 mole HC2H3O2
1 mole NaHCO3
\(\color{blue}{\text{End of Quote}}\)
*\(\color{red}{0.01}\) mole HC2H3O2
\(\color{red}{0.01}\) mole NaHCO3

Answer 18

what about 4 and 5 did i do those right?

Answer 19

1000 ml = 1 L = 1.61 moles acetic acid = 48 g acetic acid 9.86 ml = 9.86/1000 x 1.61 moles acetic acid = .0158746 moles acetic acid = .76 g acetic acid ..so 4. is correct

Answer 20

and is 5 correct?

Answer 21

@aaronq

Answer 22

@Zale101

Answer 23

Can we maybe start a new thread? It's not fun when 6 questions are posted, its very confusing.

Answer 24

its only 1 question

Answer 25

its at the top

Answer 26

can you re post it here, I'm on my phone and its really confusing.

Answer 27

5) calculate the percent of acetic acid in the vingar( the density of vingar is 1.002g/mL)
*im not really sure how to do this

Answer 28

you aren't given any mass of the acetic acid?

Answer 29

what?

Answer 30

Like, are you given grams of acetic acid?

Answer 31

it was from a lab i had to do and these are the questions but i dont know how to do this one

Answer 32

would it help if i put up the data table i did?

Answer 33

Well if you have grams, you would just convert it to moles and use the molarity to find the percent. Otherwise you would find the molar mass of the acid and divide it by the density. Either way works.


https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=molar%20mass%20of%20acetic%20acid

Molar mass of acid: 60.05g

60.06g/density

60.05g/1.002ml =

X100 for percentage =

Answer 34

60.05grams ****

Answer 35

so would it be 60%

Answer 36

rounded, yes. :D

Answer 37

thank you for your help :)

Answer 38

No problem :) good night :D