2020/2021 WAEC/NECO Answers: 2020 neco Chemistry practical answers



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 Paper III & II: Objective & Essay Government 2hrs 40mins 2:00pm 4:40pm


Government OBJ:
1-10: BECCACEDCB
11-20: CEBDDEABCC
21-30: DBAEEAEDAA
31-40: CACBAEDBAD
41-50: CCECCADCCB
51-60: BAACBDBBAA


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M
ake sure you use your school titre value

For example, if your school uses 23.50

Our own solution centre may use 24.00

Pls don't be confused, just make adjustment from the beginning where you will need to find the *VOLUME. OF ACID USED(CA)*



EXAMPLES FOR 22.50cm³

Volume. Of acid used= 1st + 2nd + 3rd/3

=24.00+24.00+24.00 / 3


= 72/3

=24 molperdm³


VOLUME. OF ACID USE for 25.00


VA= 25.00+25.00+25.00 DIVIDED BY 3

=75/3

===================================

(1)
The volume or pipette used 25.00cm³

In a tabular form
Titrations     |rough |1st|2nd|3rd|
Final reading  |21.00|41.30|20.40|40.90
Initial reading|0.00|21.00|0.00|20.40
Vol of A used  |21.00|20.30|20.40|20.50

(1ai)
Average volume of A = 20.30+20.40+20.50/3
=61.2/3
=20.40cm³

(1aii)
H2X(aq) + 2NaOH(aq)-->Na2X(aq) + 2H2O(l)

(1bi)
Mass conc. of B = 1g/250cm³
= 1g/0.25dm³ = 4g/dm³
Using :
Molar conc. = mass conc./molar mass
= 4/(23+16+1) = 4/40 = 0.1mol/dm³

(1bii)
Using CAVA/CBVB = nA/nB
CA = CBVBnA/VAnB
CA = 0.1×25×1/20.4×2
Concentration of A = 0.06127mol/dm³

1C

(i)it might contain traces of impurities which must be removed before being weighed
(ii)it is hygroscopic which readily absorb moisture from atmosphere

==========================================
NO2

1st line:|test|:Xn +distilled water,|observat
ion|:Xn dissolves in distilled water,|inference|:Xn is a soluble salt
= = = = = = = = = == = = = =
2nd line:|test|:1st portion+NAOH in drops and in excess,|observation|:pale blue ppt,|inference|:CU2+
= = = = = = = = == = = == =
3rd line:|test|:2nd portion+NH3 in drop and in excess,|observation|:deep blue ppt,|inference|
:CU2+
= == = = = = = == == = = = =
4th line:|test|3rd portion+BACL2,|
observation|:white ppt,|inference|
:SO42-,SO32-,CO32-,CL-
= = = = = = === = == = = == =
5th line:|test|:+ dil HCL,|observation|:insoluble in HCL,|Inference|:SO42-

Still the same


(2)
FILL IN THE TABLE:

(2i)
OBSERVATION:
Solution turns blue

INFERENCE:
Copper salt suspected.

(2ii)
OBSERVATION:
Blue gelatinous precipitate

Insoluble in excess NaOH

INFERENCE:
Cu²+ suspected

Cu²+ Confirmed

(2iii)
OBSERVATION:
Blue gelatinous precipitate

Soluble in excess NH³

INFERENCE:
Cu²+ Suspected

Cu²+ Confirmed

(2iv)
OBSERVATION:
White precipitate

INFERENCE:
SO4²- Suspected

(2v)
OBSERVATION:
Precipitate insoluble in excess

INFERENCE:
SO4²- confirmed

==≠=============≠===================

(3ai)
phenolphthalein

(3aii)
I. Pink
ll. Colourless

(3aiii)
(i)To give a sharp change in colour at the end of the acid base reaction

(3aiv)
(i)the burette must be rinsed with the acid or allow it to drain after rinsing with distilled water
(ii)Air bubble must not be allowed in the burette

(3av)
Neutralization reaction

(3bi)
I. Hofmann voltameter.
II. Kipp's apparatus

(3bii)

Because silver nitrate is photosensitive (i.e.sensitive to light) Therefore,if the silver nitrate is left in sunlight or any bright light, it breaks down (hydrolyses) to give black/brown silver oxide and nitric acid.




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waec expo runs is real. WAEC Answers, NECO ANSWERS and GCE answers comes a night before exam or Hours before exam

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Under the new WAEC grading system, A1 is Excellent, B2 Very Good, B3 Good, C4, C5 and C6 are interpreted as Credit, D7 and D8 are interpreted as Pass, while F9 is Fail. A1 and B2 in the WASSCE means Excellent, B3 is B (Very Good), C4 is C (Good), C5 and C6 are D (Credit), D7 and E8 are E (Pass) and F9 is F (Fail).


waec expo runs is real. WAEC Answers, NECO ANSWERS and GCE answers comes a night before exam or Hours before exam

How is Waec aggregate calculated?

A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.


A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.
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