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# 2020 Neco Chemistry Prac answes

(1)
Volume of pipette is 25.00cm³

Titrations |Rough| 1 | 2 | 3
Vol. Of acid used|23.00|21.50|21.40|21.50

(1a)
Vol of acid used VA = 21.50+21.40+21.50/3
= 21.47cm³

(1bi)
Conc. Of A in g/dm³ = 3.65÷500/1000 = 7.3g/dm³
Conc. of A in mol/dm³ = gram conc/molar mass
= 7.3/[1+35.5] = 7.3/36.5
=0.2mol/dm³

(1bii)
Using CAVA/CBVB = nA/nB
0.2×21.47/CB×25.00 = 2/1
CB = 0.2×21.47/25.00×2
= 0.08588mol/dm³

(1biii)
Molar mass of B = gram conc/molarity = 10.60/0.08588
=123.43g/mol

(1biv)
X2CO3 = 123
2[X] + 12 + 3 = 123
2x + 60 = 123
2x = 123 – 60 = 63
X = 63/2 = 31.5

(1bV)
No of moles of X2CO3 = molarity × volume
= 0.08588 × 25
= 2.147moles

Mole ratio of X2CO3 to CO2 is 1:1
Therefore, No of moles of CO2 released = 2.147moles
Volume of CO2 released = 2.147 × 22.4
=48.1dm³

(2ai)
Tabulate:
TEST: C + 5cm³ of distilled water and shake thoroughly. Divide the solutions into three portions.

OBSERVATION: A pale green solution results

INFERENCE: Salt is soluble

(2aii)
TEST: To the first portion add NaOH solution in drys

OBSERVATION: Dark green gelatinous was formed

INFERENCE: Fe²+ present

(2aiii)
TEST: then in excess

OBSERVATION: Precipitate is insoluble

INFERENCE: Fe²+ present

(2aiv)
TEST: To the second portion add K3Fe(CN)6 solution

OBSERVATION: A dark blue precipitate formed

INFERENCE: Fe²+ confirmed

(2bi)
TEST: To the third portion add AgNO³ solution

OBSERVATION: White Precipitate formed

INFERENCE: SO4²- , CL- CO²- present

(2bii)
TEST: To the results obtained in add dilute HNO3 in drops , then excess

OBSERVATION: White Precipitate is insoluble

INFERENCE: CL- present

(2biii)
TEST: To the results obtained in add NH3 solution

OBSERVATION: White Precipitate dissolves

INFERENCE: CL- Confirmed

(3ai)
Filtration

(3aii)
Diagram

(3bi)
I. Methyl Orange
II. Methyl orange
III. Phenolphthalein

(3bii)
A white precipitate soluble in excess dilute HCl is formed

Posted by on October 21, 2020.

Categories: Neco

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