September 16, 2015

GET Waec Gce 2015/2016 Nov/Dec Chemistry 3 (Alternative to Practical Work) Questions & Answers/Now Available Here

WEDNESDAY 16TH SEPT 2015
 Chemistry 3 (Alternative to Practical Work) 
9.30am - 10.00 am - 


1a)
i)
Mass concentration of B=3.95/750*1000
=5.2g/dm
VA=22.70cm^3
VB=25.0cm^3
ii)
Concentration of B in mol/dm^3
=Concentration in g/dm^3/Molar mass in g/mol
Molar mass of Na2CO3
=(23*2)+(12)+(16*3)
=46+12+48
=106g/mol
Concentration in mol/dm^3=5.27/106
=0.0497mol/dm^3
iii)
CAVA/CBVB=nA/nB
(CA*22.70)/(0.0497*25)=1/1
CA=(0.0497*25*1)/22.70*1)
CA=0.0548mol/dm^3
iv)
CA=concentration in g/dm/molar mass
Molar mass of A=H2SO4
=(1*2)+32+(16*4)
=2+32+64
=98g/mol
0.0548=concentration in g/dm^3/98
Concentration of A in g/dm^3=0.0548*98
=5.37g/dm^3
1b)
0.0548=mole/1
mole=0.0548mol
0.0548=x/6.02*10^23
x=0.0548*6.02*10^23
x=3.3*10^22
1c)
Methyl orange;This is because it is between strong acid and
weak base

2)
a)Inference:-C is a soluble salt
bi)Observation:-A gas with irritating smell is formed
bii)Observation:-A dense white fume of NH4Cl is formed
c)Observation:-Soluble in excess
d)Observation:-A gas that turns lime water milky and red
litmus paper to blue
ei)Inference:-Ca^2a+ -present
eii)Observation:-White precipitate is formed,insoluble
fi)Filtrate->CaSO4
fii)residue->CaCO3
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
3i)
Q->H2
R->CuO
X->H2O
Y->Cu
3ii)
Blue to Black
3iii)
X is serves as cooling
3b)
-Ammonia
-Hydrogen sulphide
-Chlorine
-Ammonia
-Bromine
3c)
-To measure a known volume of gas
-To keep clean substance
-To deliver a substance
-To keep a substance dry
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
1a)
i)
Mass concentration of B=3.95/750*1000
=5.2g/dm
VA=22.70cm^3
VB=25.0cm^3
ii)
Concentration of B in mol/dm^3
=Concentration in g/dm^3/Molar mass in g/mol
Molar mass of Na2CO3
=(23*2)+(12)+(16*3)
=46+12+48
=106g/mol
Concentration in mol/dm^3=5.27/106
=0.0497mol/dm^3
iii)
CAVA/CBVB=nA/nB
(CA*22.70)/(0.0497*25)=1/1
CA=(0.0497*25*1)/22.70*1)
CA=0.0548mol/dm^3
iv)
CA=concentration in g/dm/molar mass
Molar mass of A=H2SO4
=(1*2)+32+(16*4)
=2+32+64
=98g/mol
0.0548=concentration in g/dm^3/98
Concentration of A in g/dm^3=0.0548*98
=5.37g/dm^3
1b)
0.0548=mole/1
mole=0.0548mol
0.0548=x/6.02*10^23
x=0.0548*6.02*10^23
x=3.3*10^22
1c)
Methyl orange;This is because it is between strong acid and
weak base

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
 DROP YOUR  COMMENT BELOW FOR MORE ANSWERS 
Share This
Previous Post
Next Post

This Post was publish by the above Author

0 comments: