3ai)
Zinctrioxonitrate (v) – Zn(NO3)2
(3aii)
2Zn(NO3)2(g) –> 2ZnO(s) + 4NO2(g) + O2(g)
(3aiii)
The residue when it has yellow colour which will turn white on cooling
(3b)
Given; M1 = 1.0mol/dm³
V1 = ?
M2 = 0.2mol/dm³
V2 = 250cm³
Using M1V1 = M2V2
1 × V1 = 0.2×250
V1 = 50cm³
Procedure: Measure out 50cm³ of the stock solution, dilute it to 0.2mol/dm³ by adding 200cm³ of water.
(3c)
Al(SO4)3 will turn blue litmus paper (1a)
Burette reading (cm)³|1st reading|2na Reading|3rd reading|
Final |15.25|30.53|45.79|
Initial |0.00|15.25|30.53|
Volume of acids used |15.25|15.28|15.26
Average volume of acid used =15.25+15.26/2
=15.255cm³
=15.26cm³
Or =15.27cm³
(1b)
Given; mass con of A =5g/500cm³ = 5g/0.5dn³
CA=10g/dm³
A is HNO3
Therefore molar mass =1+14+(16*3)=15+48=63g/mol¹
Molarity of A = gram con/molar mass
CA=10/63=0.1587mol/dm3
(1bii)
Using CAVA/CBVB =Na/NB
With reacting equation ; HNO³ + NaOH—NaNO3+H2O
nA=1 nB=1
0.158715.26/CB25.00=1/1
25CB=0.1587*15.26
CB=0.1587*15.26/25
CB=0.09687mol/dm³
(1biii)
B is NaOH
Molar mass
23+16+1=40g/mol
Conc of B in g/dm³=molarity * molar mass
= 0.09687*40
=3.8748g/dm³
(1biv)
No of moles present in 250cm³ of NaOH is =molar conc * volume
=0.09687*250/1000
=0.0242moles
Mole ratio of NaOH and NaNO3 is 1;1
No; of mole of NaNO3 which reacted is 0.0242
Mass of NaNO3 formed =molar mass * no; of moles
=85*0.0242
=2.05grams
2a)
TEST: C+burning splint
OBSERVATION: Sample C burst into flame .It burns with non smoky blue flame without soot. Colorless gas that turns wet blue litmus paper faint red and turns like water milky is present.
INFERENCE: C is volatile and flammable. The gas is CO2 from combustion of saturated organic compound.
(2bi)
TEST: C + distilled water + shake
OBSERVATION: Clear or colorless solution is observed
INFERENCE: C is miscible with water
(2bii)
TEST: C + Acidified K2Cr207
OBSERVATION: Orange color of K2Cr207 solution turns pale green and eventually pale blue on cooling
INFERENCE: C is a reducing agent
(2d)
Compound belongs to the class of secondary alkanol
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(1a)
In a tabular form
Burette Reading|1st reading|2nd reading|3rd reading
Final |15.25|0.00|45.79|
Initial|0.00||15.25|30.53|
Volume of acid used |15.25|15.28|15.26
Average volume of acid used = 15.25 + 15.26/2
= 15.255cm³
=15.26cm³
(1bi)
Given: Mass of conc of A = 5g/500cm³ = 5g/0.5dm³
Ca = 10g/dm³
A is HNO3
Therefore; Molar mass = 1+14+(16×3)
= 15+48
=63g/mol
Molarity of A = gram conc/molar mass
Ca = 10/63 = 0.1587mol/dm³
(ii) Using CaVa/CbVb = nA/nB
With reacting equation:
HNO3 +NaOH–>NaNO3 + H2O
nA = 1, nB = 1
0.1587×15.26/Cb×25.00 = 1/1
25Cb = 0.1587×15.26
CB = 0.1587×15.26/25
CB = 0.09687mol/dm³
