serenityaliyu:
heard it will commence by Feb 29, not sure tho pls post those questions icecalm

okay, probably I will try to post it latest the night of Feb 28th

icecalm:
sorry, I didn't get the mention earlier.

no qualms sir. I hope you'll be around henceforth. Just a few days to go.

icecalm:
okay, probably I will try to post it latest the night of Feb 28th

exam starts on the 27th. It's no longer 29th. In fact, my name has been drafted among those sitting for the exam on the 27th.

Abeg, I will be the one posting the questions. That is, the ones I couldn't solve from the past questions.

Geofavor:

exam starts on the 27th. It's no longer 29th. In fact, my name has been drafted among those sitting for the exam on the 27th.

Abeg, I will be the one posting the questions. That is, the ones I couldn't solve from the past questions.

icecalm:
okay, probably I will try to post it latest the night of Feb 28th

exam starts on the 27th. It's no longer 29th. In fact, my name has been drafted among those sitting for the exam on the 27th.

Abeg, I will be the one posting the questions. That is, the ones I couldn't solve from the past questions.

I advice others too to focus on practising the past questions now.

ShyNeptune:

.... It is well.

Geofavor:

.... It is well.

Lol ... as your name is in this First batch so, that is how it will be in the first batch admission list of your desired institution. .... AmeN!

icecalm:
okay, probably I will try to post it latest the night of Feb 28th

No case.

1) The presence of an impurity in substance will cause the melting point to _____ (A) be zero. (B) reduce. (C) increase. (D) b stable

2) three drops of a 1.0 mol/dm³ solution of HCl was added to 20cm³ of a solution of pH6.4. The pH of the resulting solution will be........ (A). Close to that of pure water. (B). Less than 6.4. (C). Greater than 6.4. (D). Unaltered


pls explain how you arrived at the answer. Thanks.


Icecalm, mathefaro, oluchidelly, shyneptune, oxytocin, orezy5

i think Number 1 should be C. It says here that it's B.

Calculate the volume in cm³ of oxygen evolved at stp when a current of 5A is passed through acidified water for 193s

shyneptune, oxytocin, mathefaro, oluchidelly, orezy5, icecalm

Geofavor:
1) The presence of an impurity in substance will cause the melting point to _____ (A) be zero. (B) reduce. (C) increase. (D) b stable

2) three drops of a 1.0 mol/dm³ solution of HCl was added to 20cm³ of a solution of pH6.4. The pH of the resulting solution will be........ (A). Close to that of pure water. (B). Less than 6.4. (C). Greater than 6.4. (D). Unaltered


pls explain how you arrived at the answer. Thanks.

i think Number 1 should be C. It says here that it's B.

yeah. number one is C. because impurities increase melting point.

number two is......... Errrm. ... 'B' I think. .... because addition of an acid to a substance will reduce the pH value of the substance .....
my thoughts though.

Geofavor:
1) The presence of an impurity in substance will cause the melting point to _____ (A) be zero. (B) reduce. (C) increase. (D) b stable

2) three drops of a 1.0 mol/dm³ solution of HCl was added to 20cm³ of a solution of pH6.4. The pH of the resulting solution will be........ (A). Close to that of pure water. (B). Less than 6.4. (C). Greater than 6.4. (D). Unaltered


pls explain how you arrived at the answer. Thanks.


Icecalm, mathefaro, oluchidelly, shyneptune, oxytocin, orezy5

i think Number 1 should be C. It says here that it's B.

Impurities elevate the boiling points of substances and depress/reduce their melting point

Geofavor:
Calculate the volume in cm³ of oxygen evolved at stp when a current of 5A is passed through acidified water for 193s

shyneptune, oxytocin, mathefaro, oluchidelly, orezy5, icecalm

4OH^---->2H₂O + O₂ + 4e^-

V= (5 x 193 x 22400)/(4 x 96500)

V=21616000/386000

V=56.0cm³

Geofavor:
Calculate the volume in cm³ of oxygen evolved at stp when a current of 5A is passed through acidified water for 193s

shyneptune, oxytocin, mathefaro, oluchidelly, orezy5, icecalm

****Modified*****

0.056dm³

Owk... Here is my solution which I think it's right

m/mm =It/F

O2 +4ê= 2O^-2

F=96500× 4

m/32 = 193×5/96500×4

m= 32×193×5/96500×4

m=0.08g

At STP

32g>>22.4dm³
0.08g>>X

X=0.08×22.4/32

X=0.056dm³

20cm³ of a gaseous hydrocarbon were mixed with 90cm³ oxygen and the mixture exploded. At room temperature, 60cm³ of a gas were left. 40cm³ of the gas [Carbon (IV) Oxide] were absorbed by sodium hydroxide, leaving 20cm³ of oxygen. Find the empirical formula of the hydrocarbon.

C.c ShyNeptune Geofavor Mathefaro Orezy5 ThankyouJesus

Orezy5:

4OH^---->2H₂O + O₂ + 4e^-
V= (5 x 193 x 22400)/(4 x 96500)
V=21616000/386000
V=56.0cm³

pls explain. How did you arrive at that equation?

Oxytocin:

****Modified*****
0.056dm³
Owk... Here is my solution which I think it's right
m/mm =It/F
O2 +4ê= 2O^-2
F=96500× 4
m/32 = 193×5/96500×4
m= 32×193×5/96500×4
m=0.08g
At STP
32g>>22.4dm³ 0.08g>>X
X=0.08×22.4/32
X=0.056dm³

seems you're correct, except that your answer is in dm3.
Btw, I need you to
Explaaaaaaaaiiiiiiinnnnnn.

Geofavor:

pls explain. How did you arrive at that equation?

Okay. To the best of my knowledge, during the electrolysis of of acidified water, the anions(SO4^2- and OH^-) migrate to the anode, where OH^- are discharged. Since two OH^- groups react to form one atom of oxygen, four OH^- groups will react to form a molecule of oxygen gas which was evolved.

Geofavor:

seems you're correct, except that your answer is in dm3.

Btw, I need you to

Explaaaaaaaaiiiiiiinnnnnn.

The ionic equation for the reaction is

O2 +4ê =2O^-2

That means
No of Faraday deposited will be =4×96500C
Since 1 Faraday =96500 Coulomb

Formulae for solving questions under electrolysis

m/mm =It /F
m=mass
mm=molar mass
I=current
t=time
F=Farad

Molar mass of 0xygen gas(O2) =(16×2g)=32g


So
m/32=193×5/96500×4

m=193×5×32/96500×4

m=30880/386000
m=0.08g

So the mass of oxygen evolved is 0.08g

Now at STP

32g of Oxygen will fill 22.4dm³ of the gas
0.08 g will now fill...

22.4×0.08/32
=0.056dm³ or 56cm³

I hope it's explanatory

Hy guys,Do anyone know if they are going to repeat some past questions this year like they did last year? Any rumours of such??

Geofavor:

exam starts on the 27th. It's no longer 29th. In fact, my name has been drafted among those sitting for the exam on the 27th.

Abeg, I will be the one posting the questions. That is, the ones I couldn't solve from the past questions.

I advice others too to focus on practising the past questions now.

that's a good idea

Oxytocin:

The ionic equation for the reaction is
O2 +4ê =2O^-2
That means No of Faraday deposited will be =4×96500C Since 1 Faraday =96500 Coulomb
Formulae for solving questions under electrolysis
m/mm =It /F m=mass mm=molar mass I=current t=time F=Farad
Molar mass of 0xygen gas(O2) =(16×2g)=32g

So m/32=193×5/96500×4
m=193×5×32/96500×4
m=30880/386000 m=0.08g
So the mass of oxygen evolved is 0.08g
Now at STP
32g of Oxygen will fill 22.4dm³ of the gas 0.08 g will now fill...
22.4×0.08/32 =0.056dm³ or 56cm³
I hope it's explanatory

it's fair enough. Thanks

Oxytocin:
20cm³ of a gaseous hydrocarbon were mixed with 90cm³ oxygen and the mixture exploded. At room temperature, 60cm³ of a gas were left. 40cm³ of the gas [Carbon (IV) Oxide] were absorbed by sodium hydroxide, leaving 20cm³ of oxygen. Find the empirical formula of the hydrocarbon.

C.c ShyNeptune Geofavor Mathefaro Orezy5 ThankyouJesus

Using Avogadro's Law and Gay Lussac's law of combining volumes, the empirical formula may be obtained as described below.
Let assume a chemical equation of combustion of an unknown hydrocarbon in unlimited oxygen as
C_xH_y + nO₂ ——> xCO₂ + y/2H₂O + rO₂


From the question the volume of the remaining oxygen is 20cm³, it implies that only 70cm³ of the supplied oxygen was used.

Let assume z=n-r (both in moles and volume)

From the chemical equation above, one mole of hydrocarbon would combine with z moles of oxygen gas to produce x moles of carbon(iv)oxide and y/2 moles of steam.

40cm³ of carbon(iv)oxide was produced by 20cm³ of the hydrocarbon it therefore imply that, the number of moles of carbon in the carbon(iv)oxide is

X mol= 40/20×1mol=2mol

And the number of carbon atoms in 2mol of carbon is 2

Relatively to the above calculation we could obtain the volume of oxygen used to produce the 40cm³ carbon(iv)oxide

C + O₂——> CO₂
1vol of O₂ will combine with one mole of carbon to produce 1vol of CO₂
Therefore the volume of oxygen used to produce the 40cm³ carbon(iv)oxide is 40cm³

Since z=70cc (cc means cubic centimeter (cm³),
then the volume of oxygen used to produce the y/2 moles of steam is
S=70cc - 40cc= 30cc

The next step is to figure out how to obtain the volume of steam that will be formed from the 30cc of oxygen and this we will do.
First, write a balance equation to represent the formation of steam from hydrogen and oxygen gas, then observe the moles of both the reactants and products as if they were the volumes. Just like

(2vol)H₂ + (1vol)O₂——> (2vol)H₂O

It obvious, from the equation that 30cc of Oxygen will combine with 60cc of hydrogen to produce 60cc of steam.

Moles of hydrogen gas in 60cc of steam is equal to the moles of the steam and will be obtained as been done for the moles of carbon

***** 20cc of the hydrocarbon yield 60cc of steam, 1mol of the hydrocarbon will produce****

(Y/2) mol= 60/20×1mol= 3mol

Then Y =2×3mol/1mol=6

Y is the number of hydrogen atoms in the hydrocarbon

The empirical formula of the hydrocarbon is therefore

C₂H₆ and this is precisely Ethane

The combustion reaction for the hydrocarbon may now be written as
C₂H₆ + 9/2O₂——> 2CO₂ + 3H₂O + O₂

icecalm:
Using Avogadro's Law and Gay Lussac's law of combining volumes, the empirical formula may be obtained as described below.
Let assume a chemical equation of combustion of an unknown hydrocarbon in unlimited oxygen as
C_xH_y + nO₂ ——> xCO₂ + y/2H₂O + rO₂


From the question the volume of the remaining oxygen is 20cm³, it implies that only 70cm³ of the supplied oxygen was used.

Let assume z=n-r (both in moles and volume)

From the chemical equation above, one mole of hydrocarbon would combine with z moles of oxygen gas to produce x moles of carbon(iv)oxide and y/2 moles of steam.

40cm³ of carbon(iv)oxide was produced by 20cm³ of the hydrocarbon it therefore imply that, the number of moles of carbon in the carbon(iv)oxide is

X mol= 40/20×1mol=2mol

And the number of carbon atoms in 2mol of carbon is 2

Relatively to the above calculation we could obtain the volume of oxygen used to produce the 40cm³ carbon(iv)oxide

C + O₂——> CO₂
1vol of O₂ will combine with one mole of carbon to produce 1vol of CO₂
Therefore the volume of oxygen used to produce the 40cm³ carbon(iv)oxide is 40cm³

Since z=70cc (cc means cubic centimeter (cm³),
then the volume of oxygen used to produce the y/2 moles of steam is
S=70cc - 40cc= 30cc

The next step is to figure out how to obtain the volume of steam that will be formed from the 30cc of oxygen and this we will do.
First, write a balance equation to represent the formation of steam from hydrogen and oxygen gas, then observe the moles of both the reactants and products as if they were the volumes. Just like

(2vol)H₂ + (1vol)O₂——> (2vol)H₂O

It obvious, from the equation that 30cc of Oxygen will combine with 60cc of hydrogen to produce 60cc of steam.

Moles of hydrogen gas in 60cc of steam is equal to the moles of the steam and will be obtained as been done for the moles of carbon

(Y/2) mol= 60/20×1mol= 3mol

Then Y =2×3mol/1mol=6

Y is the number of hydrogen atoms in the hydrocarbon

The empirical formula of the hydrocarbon is therefore

C₂H₆ and this is precisely Ethane

The combustion reaction for the hydrocarbon may now be written as
C₂H₆ + 9/2O₂——> 2CO₂ + 3H₂O + O₂

RESPECT!

Thanks a lot bro..
Finding it a bit hard to comprehend tho

Chai, jus saw dis thread but am almost done with ur past convo.... Please is it too late?

Oxytocin:



RESPECT!

Thanks a lot bro..
Finding it a bit hard to comprehend tho

you are welcome bro, pls note that the calculation is based on stp and steam as combustion products will be water at stp.

icecalm:
you are welcome bro, pls note that the calculation is based on stp and steam as combustion products will be water at stp.

Owkk...
Noted!

From XNH₃(g) + YO₂(g) ----> ZNO(g), the value of Z is

A) 4

B) 7

C) 6

D) 5

a quite confusing question to me. Pls can you help me out here

icecalm, mathefaro, oxytocin, oluchidelly

.

.

.

Geofavor:
From XNH₃(g) + YO₂(g) ----> ZNO(g), the value of Z is

A) 4

B) 7

C) 6

D) 5

a quite confusing question to me. Pls can you help me out here

icecalm, mathefaro, oxytocin, oluchidelly

the question is not well framed and infinite number of answers do exit for Z for the values you choose for X and Y. From the options (A) 4 is an answer to the question when X=4 and Y=5