Question

#1
the final burret reading mL is 31.12
initial buret reading mL IS 15.72
volume of titrant mL is 15.49
Moles of NaOH used is 1.568*10^-3 mol
moles of HC2H3O2 neutralized by the NaOH is 1.568*10^-3 mole
molarity of dil HC2H3O2 is .0784 M

#2
final buret reading is 15.53
initial buret reading mL is 0.00
volume of titrant mL is 15.53
Moles of NaOH used is 1.57*10^-3
Moles of HC2H302 neutralized by the NaOH is 1.57*10^-3
molarity of dil HC2H3O2 is .0785 M

Answer 1

#3
the final burret reading mL is 31.12
initial buret reading mL 15.53
Volume of titrant mL 15.59
Moles of NaOH used is 1.578*10^-3
Moles of HC2H302 neutralized by the NaOH is 1.578*10^-3
molarity of dil HC2H3O2 is .0789 M

#4
the final burret reading mL is 46.73
initial buret reading mL is 31.12
Volume of titrant mL 15.61
Moles of NaOH used is 1.579*10^-3
Moles of HC2H302 neutralized by the NaOH is 1.579*10^-3
Molarity of dil HC2H3O2 is .07895 M

What is the average molarity of diluted vinegar

Answer 2

Wouldn't you just add up the molarity of dil HC2H3O2 and divide that number by 4 to get the average?

Answer 3

is is .0784+.0785+.0789+.07895/4

Answer 4

so the average molarity of diluted vinegar is .0786875

Answer 5

yes, that's what i got

Answer 6

ok now with that information can you help me. I have to find the deviation di

Answer 7

I don't know how to find deviation sorry :(

Answer 8

Nope. idk.

Answer 9

I have to find the deviation di

Answer 10

I don't know how sorry

Answer 11

@sunnycali

Answer 12

standard deviation s = sqrt ( (sum ((x - mean)^2) / (n - 1))

Answer 13

Create a table:
x (x-mean) (x-mean)^2
.0784
.0785
.0789
.07895

Add up the last column and plug it into the standard deviation formula mentioned earlier. Here n = 4.

Answer 14

.0786875

Answer 15

wait how did we find the moles of HC2H302