Use your school end point in number 1. ours is 12.50cm3.
(1a)
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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
=(23*2)+(32*2)+(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
= (127*2)*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
(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.
(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
thank you
428
Nice work
abbakar
532
Please gidifanc. Help us expo the answer
please I need help
Dominic