Question

A sample of a polycarboxylic acid W (Mr = 134) of mass 1.97 g was dissolved in water and
resulting solution titrated with 1.00 mol dm-NaOH.
(iv) c
29.4 cm3 of 1.00 mol dm-3 NaOH were required for complete neutralisation.
(v) C
(d) Use these data to deduce the number of carboxylic acid groups present in one molecule
W.
moles of NaoH = 1x29.4
looo
= 0.0294 moles
coboalie
acid
1.98. 0.0147
134
males Ratio - 60294: 0.0147 = 2:1
so, w contains 2 carboxylic acid groups
C
1.​

Answer 1

Explanation:

s = d ÷ t. speed = distance ÷ time.

a = (v-u) ÷ t. acceleration = change in velocity ÷ time.

F = m x a. Force = mass x acceleration.

w = m x g. weight = mass x gravity.

p = m x v. momentum = mass x velocity.

(mv - mu) = F x t. change in momentum = Force x time.

d = m v. density = mass ÷ volume.

Answer 2

Answer:

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