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Saturday, May 11, 2019

πŸ’―% CORRECT 2019 WAEC CHEMISTRY PRACTICAL ANSWERS HERE!

WAEC 2019- CHEMISTRY PRACTICAL

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Each school is to use their own end point
Many reading will be dropped
It will be better you choose to one you think or know it’s in line with your school reading because if yours is different from others rang of end point then you have failed your titration
Please take noteπŸ’―
WARNING⁉WARNING⁉
CHEMISTRY PRACTICAL
All CANDIDATES should use their school average titre value in 1a cos each school have their general titre value for their students which should differ a little with their fellow classmates due to errors encountered during titration in the lab. So when they see 12.50cm3, they should check the range they got during the titration and use it to edit our solution. Some school got 13.00cm3, 12.90cm3, 15.50cm3, etc. They should edit and recalculate the 1bii by editing our 12.50 there with their value. But if their school, got ranges like 12.30cm3-12.70cm3, they can use 12.50cm3. It’s difference is always +- 0.2.
Thanks
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(1a)
DRAW A TABLE WITH THE FOLLOWING
Burette reading|ROUGH|1st |2nd |3rd
Final |13.60|15.50|14.80|12.50
Initial |1.00 |3.10 |2.40 |0.00
Vol. acid used |12.60|12.60|12.40|12.50
Volume of S2O3²- =12.60+12.40+12.50/3
=37.50/3
=12.50cm³
(1bi)
Conc. in g/dm³ = molar mass * conc in mol/dm³
But molar mass of Na2S2O3
=(232)+(322)+(16*3)
=46 + 64 + 48
=158g/mol
Therefore 15.8g/dm³ = 158g/mol * conc of A in mol/dm³
Conc of A in mol/dm³ = 15.8/158
=0.1mol/dm³
(1bii)
Using CaVa/CbVb = na/nb
0.1*12.50/Cb * 25.00 = 2/1
50Cb = 1.25
Cb = 1.25/50
Cb = 0.025mol/dm³
The conc of I2 in B = 0.025mol/dm³
(1biii)
Gram conc of I2 = molar mass * molar conc
= (1272)0.025
=6.35g/dm³
Percentage mass of I2 in sample = 6.35/9.0 *100%
= 0.7056 * 100%
=70.56%
(1c)
There have to be a change in the colour of the mixture before it is added. The end point is known when the blue colour formed as starch is added, changes to colourless
(2a)
Test: C + Water and filtered
Observation: part of C dissolves, colour less filtrate is formed
Inference: C is a mixture of a soluble salt
(2bi)
Test: 2cm^2 of filterate of salt C + AgNO3(aq) then dilute HNO3
Observation: white precipitate is formed
Inference: Cl^- present
(2bii)
Test: Solution im 2(b)(i) + excess NH3 solution
Observation: precipitate disorder in excess solution of NH3
Inference; Cl^- confirmed
(2ci)
Test: Resident + dilute HCl + shake
Observation; residue dissolved to form a blue solution
Inference; Cu^2+ present
(2cii)
Test; solution in 2(c)(i) + NH3 solution in drops and then in excess
Observation: light pale blue precipitate is formed precipitate dissolved in excess solution of NH3 to give a deep blue solution
Inference; Cu^2+ confirmed.
(3ai)
On adding BaCl2 solutions to a portion of saturated Na2CO3 precipitate is formed,precipitate dissolves on adding excess dilute HCl
(3aii)
Q is reducing agent like SO2, H2S, CO
(3bi)
calcium oxide(Cao)
(3bii)
concentrated H2SO4
(3c)
NaOH pellet is delinquescent because it absorb moisture from the atmosphere to form solution

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