3ai)
Drawing lebel
-lid
-sticky tape(to hold paper)
-choroatographic paper
-gas jar
-applied spot of solute mixture
-ion
3aii)
principle of dillution, C1 V1=C2 V2
2.0molldm^3*40
3bi)
ethanol---->ethanoic (acid)---->trioxonotrate
(v) (acid)
3bii)
indicators are used to determine the end point
of the reaction and to test the acidity or
alkalinity of a solution
(2ai)
Observation: The salt sample dissolves completely
Inference: C is a mixture of soluble salts
(2aii)
Observation: White precipitate insoluble in excess NaOH
(aq)
Inference: Ca^2+ present
(2aiii)
Observation: No visible reaction
Inference: Ca^2+ is present
(2aiv)
Observation: Brownish fume is produced
Inference: NO2(g) form, NO3^- present
(2av)
Observation: Formation of brown gas which forms a ring
junction with FeSO4
Inference: NO2 form, NO3^- confirmed
(2bi)
Observation: D dissolves completely. Formation of dirty
green precipitate insoluble in excess NaOH(aq)
Inference: D is soluble in salt, Fe^2+ present
(2bii)
Observation: Formation of a white precipitate.precipitate
remains insoluble in HCl(aq)
Inference: SO3^2-,CO3^2-,SO4^2- present,SO4^2-
confirmed
1)
volume of pipette used =25.0cm^3
indicator used =methyl orange
burette reading|rough| 21.50|0.00|21.50|
final burette readings (cm^3)|1st|21.20| 0.00. 21.20|
initial burette reading(cm^3)|2nd|31.20|10.00| 21.20|
volume of a used (cm^3)|3rd|41.20|20.00|21.20|
average volume of a used =(21.20+21.20+21.20)cm^3/3
=63.60cm^3/3
=21.20cm^3
equation of the reaction
H2SO4(aq) + 2NaoH(aq)+2H2O(s)
nA=1,nB=2
A=4.90gldm^3 H2SO4
molar mass H2SO4
=(1*2)+32+(16*4)
=2+32+64
=98gmol^-1
molar conc A=mass con A/molar mass A
=4.90gldm^3/98glmol
=0.05moldm^3
CAVA/CBVB=nA/nB
0.05moldm^3*21.20cm^3/CB*25.0cm^3=1/2
0.05moldm^3*21.20*2=CB*25.0
CB=0.05molldm^3*21.20*2/25.0
CB=0.0848moldm^3
in mass conc B =Molar B*MolarBmass
B=NaoH
=23+16+1
=40gmol-^3
mass conc B =0.0848moldm^3 * 40gmol
mass con B = 3.392gldm^3
% mass of impusx 100%/total mass of implusty
mass of implus =(4.0-3.392)gldm^3
=0.608gldm^3
%impurity =0.608gldm^3*100%/4.0gldm^3
=15.2%
(3ai)
S is hydrogen sulphate(H2S),
T is Iron (III) sulphide V is sulphur(IV) oxide
(3aii)
It decolorizes the Solution
(3bi)
Na2CO3 is soluble in water while
NaHCO3 is insoluble in water
Na2CO3 is a basic salt which changes red
litmus paper to blu while NahCO3 is a
acidic salt that changes blue litmus
paper to read.
(3bii)
Na2CO3 cannot be decomposed
on heating while NaHCO3 can be
decomposed on heating
2NaHCO3(g)=>Na2CO3(aq)+H20+CO2(g)
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