I've having trouble finding the average molar mass number and number of acidic hydrogens Trial 4(partner)Trial 3Sample MassTrial 1Trial 2(partner)#1374o.1978018lE0.lalNaOH BuretReadingsTrial 1Trial 2Trial 3Trial 435.10 mLmL37.20 mLFinalmL6.013.29"31.81mLRq.9 ml 31.9 mlmLInitial1.02 mlmLmL31.19.Volume DeliveredmLVolume NaOH (L) 10.0318L0.0299L 0.63119LMoles NaOH Used Bol8Ixlole 2.99x0mole 3.119xlomolemole At the equivalence point, moles H = moles OH = moles NaOHMoles H In SamplemolemolevlemoleMolar MassmolemolenolemoleAverage Molar Mass:g/moleUnknown Number:Number of Acidic Hydrogens:COPShow your calculations below. Attach additional page(s) if necessary.

Answered: Image /qna-images/answer/2248deaa-7518-4469-b50e-661445b003a5.jpg

Trial Sample Mass Trial 1 Trial 2 (partner) (partner) #1374 o.1478018 0-lal NaOH Buret Readings Trial 1 Trial 2 Trial 3 Trial 4 35.10 mL 37.20mL Final mL 6.01 3.29 31.81 mL 29.9 ml 31.19. lo0 ml 6-o Initial ml mL Volume Delivered mL mL Volume NaOH (L) 00318)10.029910.631191 Moles NaOH Used Bol8Xole 2.99x mole 3.119x/0mole mo

Trial Sample Mass Trial 1 Trial 2 (partner) (partner) #1374 o.1478018 0-lal NaOH Buret Readings Trial 1 Trial 2 Trial 3 Trial 4 35.10 mL 37.20mL Final mL 6.01 3.29 31.81 mL 29.9 ml 31.19. lo0 ml 6-o Initial ml mL Volume Delivered mL mL Volume NaOH (L) 00318)10.029910.631191 Moles NaOH Used Bol8Xole 2.99x mole 3.119x/0mole mo

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter5: Resonance

Section: Chapter Questions

Problem 18E

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I want both handwritten correct, handwritten only if u can't i suggest u to skip ;)
Ok this looks similuar to mole calculation I think but I havent seen one like this before.
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Please help, will provide helpful ratings for correct soln. Thanku
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Please solve both parts qs, will provide helpful ratings.
how can i solve the questio below? it's past paper with no markscheme so it's fine