Trial 2 Trial 3 Average(Trials 2 & 3)
Volume of NaOH used (ml) 33.00mL- 31.15mL=2.18mL 30.00mL-28.15mL = 1.85mL 2.18mL + 1.85mL /2= 2.015mL
Volume of NaOH (L): 2.18*0.001=0.002 L 1.85*0.001=0.002 L 0.002+0.002/2=0.002 L

QUESTIONS I NEED ANSWERED
Moles of NaOH (molarity x volume):
Moles of Aceitc Acid:
Volume of Acetic Acid (L):
Molarity of Acetic Acid:

Question

Trial 2	  Trial 3       Average(Trials 2 & 3)      
Volume of NaOH used (ml)        33.00mL- 31.15mL=2.18mL   30.00mL-28.15mL = 1.85mL   2.18mL + 1.85mL /2= 2.015mL
Volume of NaOH (L):                 2.18*0.001=0.002 L            1.85*0.001=0.002 L           0.002+0.002/2=0.002 L

QUESTIONS I NEED ANSWERED
Moles of NaOH (molarity x volume):
Moles of Aceitc Acid:
Volume of Acetic Acid (L):
Molarity of Acetic Acid:

anonymous · Answer

Can you post a picture of your work? I have no idea what your original post is trying to communicate.

anonymous · Answer

yes like @saguanau said please attach picture or take screenshot and than attach screenshot...

anonymous · Answer

@Kryten @saguanau I ATTACHED MY LAB! i NEED HELP ON ALL OF THE BLANK QUESTIONS. thank you very much!

anonymous · Answer

oh god you took picture of your computer screen!!! you have a key on keyboard "PrtSc" and you open your paper and push that button and then go to paint and just press ctrl+v or go to edit and push paste and save it as picture... but ill try to read this...

anonymous · Answer

@breezyxo You gotta screenshot the second picture ha, there's no way that's legible

anonymous · Answer

I attached the file this time lol. @saguanau @Kryten

anonymous · Answer

ok what is concentration of NaOH or Benzoic acid or acetic acid?

anonymous · Answer

idk... i really just need the 4 answers at the bottom

anonymous · Answer

yes but you cant calculate them without concentration, dont tell me you forgot to write down concentration of NaOH from buirette

anonymous · Answer

sorry...

anonymous · Answer

i will try and figure it out then tag you in it tomorrow