Nabteb 2015/2016 Chemistry Practicals

Nabteb Chemistry Practical Answers
3 a .
i . kipps apparatus : it is used in intermittant
supply o gasses.
ii . it is used in cooling or condersing vapour into
liquid during distilation lebigesndenses
iii . hoffmanis voltameter : i is used oin
determination of the chemical composition of
water by volume .
3 b .
i . add three or four drops of concentrated
trioxonitrate ( v ) acide of egg- white solution the
firmation of an intense yellow colour indicate the
presence of protein.
ii . crystallization
iii . when water is added to dinhydrous copper ( ii )
trioxosulphate ( vi) to the colour will change from
white to blue .
2 a . Tabulate Test, Observatiob, Inference
i .
UNDER TEST PUT :
G + 5 cm 3 of distilled water
G solution + litmuss paper
UNDER OBSERVATION PUT :
G is soluble in water
No effect in litmus paper
UNDER INFERENCE PUT :
A soluble salt is present
The soft in neutral to litmus paper.
ii .
UNDER TEST PUT :
G solution + oe or two drop of iondine
UNDER OBSERVATION PUT :
The soltuion turns blue black
UNDER INFERENCE PUT :
Glucose is present
iii .
UNDER TEST PUT :
G solution + a few drop of feliling solution and
warn
UNDER OBSERVATION PUT :
A brick red pricipitate is formed
UNDER INFERENCE PUT :
Glucose confirm.
2 b . Tabulate Test, Observatiob, Inference
i .
UNDER TEST PUT :
H + 5 cm 3 of dilute tetraoxosulphate acide and
heat
UNDER OBSERVATION PUT :
a blue gelate haay precipitate is formed
UNDER INFERENCE PUT :
CU 2 + is present
ii .
UNDER TEST PUT :
2 cm 3 b ( i ) + NH ( A) in drop then in excess
UNDER OBSERVATION PUT :
a deep blue precipitate is formed and it dissolved
in excess
UNDER INFERENCE PUT :
CU 2 + confirm
1 a .
Volume of pipette used = 25 cm 3
Tabulate: – – – – , 1 st , 2 nd , 3 rd
UNDER – – – – PUT :
Burette Reading ( cm 3 )
Initial burette reading ( cm 3 )
Volume of acid used
UNDER 1 ST PUT :
23 . 50
00 . 00
23 . 50
UNDER 2 nd PUT :
47 . 10
23 . 50
23 . 60
UNDER 3 rd Put :
23 . 40
00 . 00
23 . 40
Average volume of E used
= 23 . 50 + 23 . 60 + 23 . 40 / 3
= 23 . 50 cm 3
HNO 3 ( aq) + NaOH( aq) – – > NaNO 3 ( aq ) + H 20 ( aq )
1 b . F contains 2 . 00 g NaOH( aq) in 5 . 00 cm 3
500 cm 3 of F = 100 cm 3
2 . 00 g of F = X
500 X = 2 x 1000
X = 2000/ 5
X = 4 gldm3 of F
conc of F in gdm 3 = 4 gdm3
molar concentration = mass concentration /
molar mass
molar mass = NaOH = 23 + 16 + 1
= 40 gldm
= 4 gldm3 / 40 gldm
molar conc of F = 0 . 1 moldm 3
Using CAVA / CBVB = nA / nB
CA x 23 . 50 / 0 . 1 x 25 = 1 / 1 – > ca x 23 . 5 = 2 . 5
CA = 2 . 5 / 23 . 50 – > CA = 0 . 1064moldm 3
Concentration of E = 0 . 1064 moldm 3
recall that mass concentrate = mass / volume
( dm 3 )
form he balanced chemical equation
1 mole of F = 1 mole of NaNO 3
500 dm 3 of F = 500 dm 3 of NaNO 3
– > volume of NaNo 3 = 500 dm 3
also 1 mole of F = 1 mope of NaNO3
0 . 1 moldm 3 of F = 0 . 1 moldm 3 NaNO 3
– > Number of mole of NaNO3 = volume x
concentration
= 500 x 0 . 1
= 50 mole
Molar Mass NaNO 3 = 23 + 14 + 3 ( 16 )
= 85 gmol
mass of NaNO3 = number of mole x molar mass
= 50 x 85
= 4250 g
– – – – – – – – – – – – GOODLUCK – – – – – – – – – – – –

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