Question

PLEASE PLEASE HELP I will give a medal and fan.
I need help on a chemistry lab i had this up before but someone gave me just the answers and i would like to know how to do the actual work

Answer 1

i would need help with 2-7

Answer 2

1. . Write the balanced equation for the reaction of acetic acid
( COOH3CH ) and sodium bicarbonate (3NaHCO). What type of
reaction(s) took place?
\[ CH_{3}COOH + NaHCO_{3} \rightarrow CH_{3}COONa + H_{2}O + CO_{2}\]
there is actually two reactions taking place. the first is to produce the sodium carbonate by double displacement, the second is a decomposition that produces water and carbon dioxide.

Answer 3

might me useful to say that acid base reactions took place as well?

Answer 4

2. Calculate the number of moles of NaHCO3 that were required to
neutralize the COOH3CH in the vinegar.
from the equation we can see that there is a ratio of one mole to one mole:
\[\frac{gramsCOOH_{3}CH}{1}*\frac{1moleCOOH_{3}CH}{?gramsCOOH_{3}CH}\]
\[*\frac{1moleNaHCO_{3}}{1moleCOOH_{3}CH}\]

Answer 5

3. Calculate the molarity of the vinegar sample. (Don’t forget to
convert mL to L.)
molarity:
\[\frac{Moles}{L}\]
find how many moles you have (you will probably have to convert to grams by using the density of vinegar):
\[\frac{vinegar(grams)}{1}*\frac{moles}{grams}\]
how much solution is the vinegar in?
convert mL to L

Answer 6

4.Calculate the number of grams of COOH3CH in the vinegar
\[moles=concentration(M)*volume(C)\]
\[\frac{moles}{1}*\frac{grams}{moles}\]

Answer 7

5.Calculate the percent of acetic acid in the vinegar. (The density
of vinegar is 1.002 g/ml.)
\[MM(molar..mass)=\frac{grams}{mole}\]
\[mass ..percent=moles(found.in.number.4)*MM\]

Answer 8

6.Is it possible for the equivalence point of a titration to not be at pH 7? Explain your answer.
The only reason it will be 7 for some titrations is that when one titrates a strong acid with a strong base. In the titration of acetic acid with NaOH the equivalence point will be about 8.3-8.7 depending on how much sodium acetate is present

Answer 9

7.. What is the molarity of a CsOH solution if 30.0 mL of the solution
is neutralized by 26.4 mL 0.25 M HBr solution?
HBr is a strong acid, ionizes completely
\[Mole = (\frac{0.250 mol}{L} )( 0.0264 L) = 0.0066 mol\]
\[HBr\rightarrow H^{+} + Br^{-}\]
0.250 M . 0.250 M 0.250 M
0.0066 mol .. 0.0066 mol


\[H^{+} + OH\rightarrow H2O \]
0.0066 mol H+ neutralizes 0.0066 mol OH^- ion

CsOH is a strong base and it dissociates completely
\[CsOH_{aq} \rightarrow Cs^{+}_{aq} + OH^{-}_{aq)}\]
0.0066 mol < 0.0066 mol

\[M =\frac{ mole}{ Volume} = \frac{0.0066 mole}{ 0.0300 L }= 0.22 M\]

Answer 10

I started trying to work this out but i have a F in the class because i haven't been paying attention so i don't know where to plug things in to these equations i was wondering if you could help me step by step? I really want to try and get a D in the class at least and if some one explained step by step then maybe i could apply the knowledge to other labs. So would you please help me? @Kkutie7

Answer 11

well if you point out certain parts i can help you plug stuff in. the more you can do by yourself he more you will benefit, but I understand the need for a better grade