Question

Molarity of standardized NaOH solution is 0.1012
final buret reading, mL IS 31.21
INITIAL buret reading, mL IS 15.72
VOLUME OF TITRANT, mL is 15.49
Moles of NaOH used is 1.568*10^-3

Find Moles of HC2H302 neutralized by the NaOH?

Answer 1

Use: \(\sf \color{blue}{M_1V_1=M_2V_2}\)

Answer 2

@nincompoop

Answer 3

?

Answer 4

@LunyMoony

what'cha need?

Answer 5

Hmmm I can try (=

Answer 6

ok give me a minute (=

Answer 7

np

Answer 8

Maybe This Will Help You:

moles of NaOH = M*V = 0.1 * 0.00413 = 0.000413 mol
moles of acetic acid = moles of NaOH = 0.000413 mol
grams of acetic acid = moles * MM = 0.000413 * 60 = 0.0248 g
%(m/v) = (g / mL) *100 = (0.0248 / 5) *100 = 4.96 %

Answer 9

OMG THIS IS SO HARD! )=

Answer 10

I think we Use: M1V1=M2V2 equation

Answer 11

@AuroraB

Answer 12

@math

Answer 13

@moderator

Answer 14

HAHA yeahhhhhhhhh i can't do this... sorry... /:

Answer 15

I'm not there yet sorry

Answer 16

I know )=

Answer 17

Molarity of standardized NaOH solution is 0.1012
final buret reading, mL IS 31.21
INITIAL buret reading, mL IS 15.72
VOLUME OF TITRANT, mL is 15.49
Moles of NaOH used is 1.568*10^-3

Find Moles of HC2H302 neutralized by the NaOH?

Answer 18

the volume of diluted vinegar used in each titration is 20 ml .02 L

Answer 19

do you need to find the molarity of the acetic acid? you have to divide the number of moles by the volume in liters. 1.568*10^-3 /0.2= 7.84 x 10^-4 M

Answer 20

NO I HAVE TO FIND moles of HC2H302 NEUTRALIZED BY THE NaOH

Answer 21

IT THE ANSWER 1.568*10^-3

Answer 22

yes that the answer and already told you how to do the calculations it is that clear now?

Answer 23

YES

Answer 24

NOW I have already ruined my lab report. how do i print out a new one that is blank

Answer 25

like is there a trick

Answer 26

i want a new sheet of my lab report

as i messed up the originla one

Answer 27

@abb0t is he right