Online tutorial in Physics, Chemistry, Maths for Jamb and pume - Education (36)

Online tutorial in Physics, Chemistry, Maths for Jamb and pume - Education (36) - Nairaland

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Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 8:56pm On Jan 05, 2014

[quote author=great.david]i belive,you forgot to tell us the medium we should use....but i'll use acidic medium..
Cl03^-1 => Cl^- ( reduction)
zn => Zn^2+ (oxidation)
Cl03^-1 + 3H + 6e^- => Cl^- + 3OH
Zn => 3Zn^2+ + 2e^-
finally: Cl03^- + 3H^+ + 3Zn => Cl^- + 30H^- + 3Zn^2+[/quote]Dear sir...u are wrong......i saw so many errors that got u a bad landing!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 9:03pm On Jan 05, 2014

aysuccess99: though i don't know it wella but i would try it according to my best of knowlege. i would use alkaline medium.
[ClO(3)]^(-1)+Zn--->Cl^(-1)+Zn^(+2)
Half Cell Equation.
[Cl(O)3]^(-1)---->Cl^(-1).....eqn 1
Zn--->Zn^(2+)
[Cl(O)3]^(-1)+H20---> Cl^(-1)+OH^(-1)
Zn--->Zn^(2+)+2e^(-1)........eqn2 (*3)
[Cl(O)3]^(-1)+3H20+6e^(-1)---> Cl^(-1)+6OH^(-1)
3Zn--->3Zn^(2+)+6e^(-1)
We then add the two equations up. we now have:
[Cl(O)3]^(-1)+3Zn+3H20+6e^/(-1)--->3Zn^(2+)+6e^/(-1)+ Cl^(-1)+6OH^(-1)
Then we have our final answer as:
[Cl(O)3]^(-1)+3Zn+3H20--->3Zn^(2+)+Cl^(-1)+6OH^(-1).
please bosses, my answer is subjected to correction. If I am wrong, please do correct me. thanks.

Though you are right working with an ALKALINE MEDIUM....When next you pass a question as these...you should find reason why you should work it over in the ACIDIC MEDIUM as DAVID did..though he was wrong!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 9:08pm On Jan 05, 2014

[quote author=Dr. Laz'Devitan]Though you are right working with an ALKALINE MEDIUM....When next you pass a question as these...you should find reason why you should work it over in the ACIDIC MEDIUM as DAVID did..though he was wrong![/quote]okay bro, thanks sir. But please sir, if you are to do it in an acidic medium, how would we go about it? thanks a lot doctor to be.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 10:14pm On Jan 05, 2014*

[quote

author=aysuccess99]okay bro, thanks sir. But please sir, if you are to do it in an acidic
medium, how would we go about it? thanks a lot doctor to be.[/quote]Got a poor airtime...i pray network dnt put me half-way!
Zn--->Zn^2(OXD)............1
ClO(3)^-1---->Cl^-1(RED).........2
To eqn(2),add a molecule of water to the right side and den balance the
hydrogen by adding a proton ion to the left side of the eqn!..Finally we
shal have
2ClO(3)^-1 + 12H^+ ---> 2Cl^- + 6H(2)O.................(2)
after dat balance the charges of both REDOX EQUATION into dia seprate
half-cell!
Zn ----> Zn^2 + 2e^-...........1
2ClO(3)^-1 + 12H^+1 + 12e^-1 --> 2Cl^-1 + 6H(2)O........2
Using 2e from that of eqn to multiply althrough that of eqn(2)...n
simultaneously use 12e of that of eqn to multiply althrough that of
eqn(1)....:...
let me represent ClO(3)^- as ''A''......H^+ As ''B'' ..........Cl^- as
''M''.......So, if i jump so many steps......we may arrive at
4A + 24B +12Zn => 4M + 12Zn^2+ + 12H(2)O....After dis,then divide through all the right and left side by 4....The final balance equation is

ClO(3)^- + 6H^+ + 3Zn=>Cl^- + 3Zn^2+ + 3H(2)O

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by greatdavid1: 10:31pm On Jan 05, 2014

[quote author=Dr. Laz'Devitan]Dear sir...u are wrong......i saw so many errors that got u a bad landing![/quote]oh!! It was an error of the first order....never to make that again...i should have used H20 and H^+ for the balancing....a slap on ma face

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by greatdavid1: 10:39pm On Jan 05, 2014

[quote author=Dr. Laz'Devitan]Got a poor airtime...i pray network dnt put me half-way!
Zn--->Zn^2(OXD)............1
ClO(3)^-1---->Cl^-1(RED).........2
To eqn(2),add a molecule of water to the right side and den balance the
hydrogen by adding a proton ion to the left side of the eqn!..Finally we shal have
2ClO(3)^-1 + 12H^+ ---> 2Cl^- + 6H(2)O.................(2)
after dat balance the charges of both REDOX EQUATION into dia seprate half-cell!
Zn ----> Zn^2 + 2e^-...........1
2ClO(3)^-1 + 12H^+1 + 12e^-1 --> 2Cl^-1 + 6H(2)O........2
Using 2e from that of eqn to multiply althrough that of eqn(2)...n simultaneously use 12e of that of eqn to multiply althrough that of eqn(1)....:...[/quote]thanks for the correction sir.......

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 10:53pm On Jan 05, 2014

[quote author=great.david]thanks for the correction sir.......[/quote]We are one.......no man is MR-I-KNOW-ALL.......am humbled sir!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 5:59am On Jan 06, 2014

[quote author=Dr. Laz'Devitan]Let me give you guys some of my tipsy questions...to grind u guys up

1)45cm^3 of a mixture of carbon( II ) and carbon ( IV ) oxide were mixed
with 45cm^3 of oxygen,and exploded.The volume of the resulting
gas,measured at the same temperature and pressure,was 70cm^3.Calculate
the volume of the carbon monoxide in the original mixture?

2)A hose directs a horizontal jet of water,moving with a velocity of 20m/s,on to a vertical wall.The cross-sectional area of the jet is 0.0005m^2.If the density of the water is 1Mg/m^3,calculate the force on the wall assuming the water is brought to rest there.[/quote][color=cyan]
these questions have no solution yet
I think the questions are wrong,

prove me wrong by providing the solution please[/color]

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 8:20am On Jan 06, 2014*

Swagalord18: [color=cyan]
these questions have no solution yet
I think the questions are wrong,

prove me wrong by providing the solution please[/color]

The questions are very correct...let me solve them
1st QST

Vol of CO + Vol of CO(2)=45cm^3
Let have it that Vol of CO(2)=xcm^3
so,that of CO=(45-x)cm^3......

Dont forget the gas laws to use are

AVOGADRO'S LAW or GAY-LUSSAC LAW of GAS COMBINATION!
From reaction,we have
2Vol of CO + 1Vol of O(2)=2Vol of CO(2)
Not forgetting that volume of oxygen=45cm^3
Before sparking:
(45-x)cm^3 of CO + 45cm^3=>---- of CO(2)

After sparking:
(45-x)cm^3 of CO + (45-x)/(2)=> (45-x)cm^
Residual gas of the Oxygen= 45-(45-x)/(2)....we now have (45+x)/(2)

Total gas left that did not combined is
[(45+x)/(2)] + x + (45-x)=70
45 + x + 90 =140
x=5.....i.e. CARBON(IV)OXIDE Volume in the mixture=5cm^3
then
CARBON(II)OXIDE=45-5
=40cm^3

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 8:40am On Jan 06, 2014

[quote author=Dr. Laz'Devitan]The questions are very correct...let me solve them
1st QST

Vol of CO + Vol of CO(2)=45cm^3
Let have it that Vol of CO(2)=xcm^3
so,that of CO=(45-x)cm^3......

Dont forget the gas laws to use are

AVOGADRO'S LAW or GAY-LUSSAC LAW of GAS COMBINATION!
From reaction,we have
2Vol of CO + 1Vol of O(2)=2Vol of CO(2)
Not forgetting that volume of oxygen=45cm^3
Before sparking:
(45-x)cm^3 of CO + 45cm^3=>---- of CO(2)

After sparking:
(45-x)cm^3 of CO + (45-x)/(2)=> (45-x)cm^
Residual gas of the Oxygen= 45-(45-x)/(2)....we now have (45+x)/(2)

Total gas left that did not combined is
[(45+x)/(2)] + x + (45-x)=70
45 + x + 90 = 70
x=5.....i.e. CARBON(IV)OXIDE Volume in the mixture=5cm^3
then
CARBON(II)OXIDE=45-5 =40cm^3[/quote]Great workings there doctor, I was about solving it when I saw your correction. Is there any question from you today? Thanks in anticipation.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 8:43am On Jan 06, 2014

Swagalord18: [color=cyan]
these questions have no solution yet
I think the questions are wrong,

prove me wrong by providing the solution please[/color]

2nd QST.....
This question is about moment of changing volume with time...not as in the normal change in velocity with time
Force=Mass * accelearation
Mass=Density(d) * Volume(V)
Acc=Vel/time
Force=Density(d) * Volume * velocit/time
Volume =Area(A) * Lenght(L)
F=density * Area * Length * Vel/time
F= density * area *(Length/time) *velocity
we can now,accept dat
Lenght/time=Velocity of the water
F=density * area * Velocity * Velocity
Force=Density * area * (Velocity)^2
Force=dav^2
Force=1000 * 0.0005 * 400

Thus,

Force of the flow=200N

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 8:54am On Jan 06, 2014

[quote author=Dr. Laz'Devitan]2nd QST.....
This question is about moment of changing volume with time...not as in the normal change in velocity with time
Force=Mass * accelearation
Mass=Density(d) * Volume(V)
Acc=Vel/time
Force=Density(d) * Volume * velocit/time
Volume =Area(A) * Lenght(L)
F=density * Area * Length * Vel/time
F= density * area *(Length/time) *velocity
we can now,accept dat
Lenght/time=Velocity of the water
F=density * area * Velocity * Velocity
Force=Density * area * (Velocity)^2
Force=dav^2
Force=1000 * 0.0005 * 400

Thus,
Force of the flow=200N[/quote]confirm!!!! nice workings sir. kudos and thumbs up for you.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by iniking: 10:33am On Jan 06, 2014

UTME candidates are encouraged to avail themselves to the Free online classroom at www.iniking.com/classroom offering quality teachings from subject matter experts. Mathematics is already loading. Other subjects coming up soon.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 10:37am On Jan 06, 2014

OMG!....this time I have to say something....Dr Laz.,I really need your real identity cos ur superb.I think you are or should be a Lecturer.Plz,lemme knw whr to hook u.

Aysuccess,I dey always feel your relentless effort towards questions posted,you'll always give it a try.I saw your posts on Maths clinic too,I guess you should be an Engineer not a medic.Chop knockle boss.

@ the rest guru's,I doff o.

Should I say I'm in?well,let's go there!!!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 11:06am On Jan 06, 2014

My ogas and bosses in the thread, I am really very sorry for derailing the thread. To all science students, medical aspirants, BDS aspirants, Pharmacy, biochem, microbiology, physiotherapy, radiography aspirants and et al that will be writing BIOLOGY in this year JAMB exam, there's a thread that's wide open for us to share our opinion, ask questions and answer them and make suggestions about the answers. This is the link: https://www.nairaland.com/1210309/biology-chemistry-tutorial-thread-jamb/23#2070672
And I pray that God will answer our prayer and crown our reading with success and admission.
Once again, I am very sorry for derailing, God Bless Us All.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 1:16pm On Jan 06, 2014

The magnification of an object 2cm tall when placed 10cm in front of a plane mirror is
A. 6.0
B. 1.0
C. 0.7
D. 0.6

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 6:01pm On Jan 06, 2014

[quote
author=FrancisTony]The magnification of an object 2cm tall when placed
10cm in front of a plane mirror is
A. 6.0
B. 1.0
C. 0.7
D. 0.6[/quote]Problems with your question
1)Plane mirrors don't magnify objects,that's y u observe that you appear
of same size in a mirror...So plane mirrors dont cause magnification
2)Plane mirror don't have focal point......so,your question is baseless!

Mirrors notable for causing MAGNIFICATION are

CONCAVE and CONVEX Mirrors

They are SPHERICAL and not PLANE!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 6:41pm On Jan 06, 2014

[quote author=Dr. Laz'Devitan]Problems with your question
1)Plane mirrors don't magnify objects,that's y u observe that you appear
of same size in a mirror...So plane mirrors dont cause magnification
2)Plane mirror don't have focal point......so,your question is baseless!

Mirrors notable for causing MAGNIFICATION are

CONCAVE and CONVEX Mirrors

They are SPHERICAL and not PLANE![/quote]Thanks! The question was confusing, that was why i decided to drop it here! But u're really IWE oo.
Hmm, talented!
What uni are u aspiring for, and what course?

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 7:13pm On Jan 06, 2014

Let's check these chemistry please.

1)If a given quantity of electricity librates 0.65g of Zn^2+,what amount of Hg^2+ would be librated by the same quantity of current?[Zn=65 and Hg=201]

2)When a solution of ammonium trioxocarbonate(iv) is added to a solution of an unknown salt,a white precipitate which is soluble in dilute hydrochloric acid but insoluble in ethanoic acid is formed.This indicates the presence of

A)Ca^2+
B)Na^+
C)Zn^2+
D)K^+

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 7:32pm On Jan 06, 2014

[quote
author=labodinho]Let's check these chemistry please.

1)If a given quantity of electricity librates 0.65g of Zn^2+,what amount
of Hg^2+ would be librated by the same quantity of current?[Zn=65 and
Hg=201]

2)When a solution of ammonium trioxocarbonate(iv) is added to a solution
of an unknown salt,a white precipitate which is soluble in dilute
hydrochloric acid but insoluble in ethanoic acid is formed.This
indicates the presence of

A)Ca^2+
B)Na^+
C)Zn^2+
D)K^+[/quote]Solution to your 1st QST
Zn^2+ + 2e^-1 => Zn
65g of Zn liberates 2e
then 0.65g of Zn lib (0.65 * 2)/65
=0.02e^-..........
Hg^2+ + 2e^- => Hg
2e^- => 201g
0.02e^- =>(201 * 0.02)/(2)
Thus,Mass of MERCURY=2.01g

For ur 2nd QST

That STOICHOMETRY involving testing for the unknown substance is perculiar to

Ca^2+..........So that's the answer!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 7:50pm On Jan 06, 2014

[quote author=Dr. Laz'Devitan]Solution to your 1st QST
Zn^2+ + 2e^-1 => Zn
65g of Zn liberates 2e
then 0.65g of Zn lib (0.65 * 2)/65
=0.02e^-..........
Hg^2+ + 2e^- => Hg
2e^- => 201g
0.02e^- =>(201 * 0.02)/(2)
Thus,Mass of MERCURY=2.01g

For ur 2nd QST

That STOICHOMETRY involving testing for the unknown substance is perculiar to

Ca^2+..........So that's the answer![/quote]What a question murderer you are....Good of you sir.Preparing another one now.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Nobody: 8:40pm On Jan 06, 2014

labodinho: What a question murderer you are....Good of you sir.Preparing another one now.

I am waiting patiently for it. DR. LEVDIVATAN, welldone boss.
@oga labodinho, thanks for all the questions that you've posted. I've learnt one or two things from your questions. Thanks.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 10:01pm On Jan 06, 2014

Here we go..

1)Sulhur(iv)oxide is a strong reducing agent in the presence of water due to the formation of
A)trioxosulphate(iv) ion
B)hydroxide ion
C)sulphur(vi)oxide
D)hydrogen sulhide

2)A metal that forms soluble trioxosulphate(iv) salt is
A)Aluminium
B)Barium
C)Potassium
D)Manganese

3)Which of the following substances is not a salt?
A)Zinc chloride
B)Aluminium oxide
C)Sodium hydrogentrioxosulphate(iv)
D)Sodium trioxocarbonate(iv)

4)If 24.4g of Lead(ii)trioxonitrate(v) were dissolved in 42g of distilled water at 20DCelcius,calculate the solubility of solute in gdm^-3
A)58.100
B)581.000
C)0.581
D)5.810

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 10:03pm On Jan 06, 2014

aysuccess99: I am waiting patiently for it. DR. LEVDIVATAN, welldone boss.
@oga labodinho, thanks for all the questions that you've posted. I've learnt one or two things from your questions. Thanks.

Ma boss Aysuccess,you're welcome sir.

@all,sorry ma question came late,was kinda busy.So let's manage Monday Dinner.

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Aibike(m): 10:41pm On Jan 06, 2014

Pls d gurus in d aus,as u take mr Labodinho's Q,U can as well break dis down 4 us!

19.04g of ammonia were mixed with 31.10g of hydrogenchloride gas in a closed container,
a.whch of the reactants was in excess and by howmuch
b.how much ammoniumchloride was formed
c.how much more of d insufficient reactant would b needed to completely react with excess of other reactant?
(N=14,H=1, Cl=35.5)

pls,I need it worked out!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Influence01: 10:42pm On Jan 06, 2014

labodinho: Here we go..

1)Sulhur(iv)oxide is a strong reducing agent in the presence of water due to the formation of
A)trioxosulphate(iv) ion
B)hydroxide ion
C)sulphur(vi)oxide
D)hydrogen sulhide

2)A metal that forms soluble trioxosulphate(iv) salt is
A)Aluminium
B)Barium
C)Potassium
D)Manganese

3)Which of the following substances is not a salt?
A)Zinc chloride
B)Aluminium oxide
C)Sodium hydrogentrioxosulphate(iv)
D)Sodium trioxocarbonate(iv)

4)If 24.4g of Lead(ii)trioxonitrate(v) were dissolved in 42g of distilled water at 20DCelcius,calculate the solubility of solute in gdm^-3
A)58.100
B)581.000
C)0.581
D)5.810

1.A (not sure)
2.C (group 1 metals like sodium, potassium and ammonia forms soluble salts)
3.B (Al2O3 is an amphoteric oxide)
4.A (not sure)
; D

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 10:47pm On Jan 06, 2014*

labodinho: Here we
go..

1)Sulhur(iv)oxide is a strong reducing agent in the presence of water
due to the formation of
A)trioxosulphate(iv) ion
B)hydroxide ion
C)sulphur(vi)oxide
D)hydrogen sulhide

2)A metal that forms soluble trioxosulphate(iv) salt is
A)Aluminium
B)Barium
C)Potassium
D)Manganese

3)Which of the following substances is not a salt?
A)Zinc chloride
B)Aluminium oxide
C)Sodium hydrogentrioxosulphate(iv)
D)Sodium trioxocarbonate(iv)

4)If 24.4g of Lead(ii)trioxonitrate(v) were dissolved in 42g of
distilled water at 20DCelcius,calculate the solubility of solute in
gdm^-3
A)58.100
B)581.000
C)0.581
D)5.810

I did not see it properly....pardon my error...this the proper answer
1)A 2)B 3)B 4)B

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 11:26pm On Jan 06, 2014*

[quote author=Ai-bi-ke]Pls d gurus in d aus,as u take mr Labodinho's Q,U can as well break dis down 4 us!

19.04g of ammonia were mixed with 31.10g of hydrogenchloride gas in a closed container,
a.whch of the reactants was in excess and by howmuch
b.how much ammoniumchloride was formed
c.how much more of d insufficient reactant would b needed to completely react with excess of other reactant?
(N=14,H=1, Cl=35.5)

pls,I need it worked out![/quote]This a matter of throwing trials to the calculation to knw which is excess or not,
1mol NH(3) + 1mol HCl=> 1mol of NH(4)Cl
Mass before sparking:
19.04g NH(3) + 31.10g HCl=>........
using
36.5g HCl => 17g NH(3)
31.10g HCl => (31.10 * 17)/(36.5)
so we have 14.48g of NH(3)
when i tried the other......I got
40.58g of HCl.....this means that HCl was just enough while the NH(3) was in excess
a) So NH(3) was in excess and by an amount of (19.04 - 14.48)=4.56g

b) so we then can use 14.48g of NH(3) or 31.10g of HCl to calculate that of the AMMONIUM ION.....
36.5g of HCl=> 53.5g of NH(4)Cl
31.10g of HCl=>(31.1 * 53.5)/(36.5)
so, the Mass amount of the ammonium ion=45.58g

c)To get the other amount to react with the excess NH(3).....You do it as follow!
17g of NH(3) => 36.5g of HCl
4.56g of NH(3) => (4.56 * 36.5)/(17)
=9.79g of HCl
So the mass just need is
9.79g of HCl..!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by Aibike(m): 11:59pm On Jan 06, 2014

[quote author=Dr. Laz'Devitan]This a matter of throwing trials to the calculation to knw which is excess or not,
1mol NH(3) + 1mol HCl=> 1mol of NH(4)Cl
Mass before sparking:
19.04g NH(3) + 31.10g HCl=>........
using
36.5g HCl => 17g NH(3)
31.10g HCl => (31.10 * 17)/(36.5)
so we have 14.48g of NH(3)
when i tried the other......I got
40.58g of HCl.....this means that HCl was just enough while the NH(3) was in excess
a) So NH(3) was in excess and by an amount of (19.04 - 14.48)=4.56g

b) so we then can use 14.48g of NH(3) or 31.10g of HCl to calculate that of the AMMONIUM ION.....
36.5g of HCl=> 53.5g of NH(4)Cl
31.10g of HCl=>(31.1 * 53.5)/(36.5)
so, the Mass amount of the ammonium ion=45.58g

c)To get the other amount to react with the excess NH(3).....You do it as follow!
17g of NH(3) => 36.5g of HCl
4.56g of NH(3) => (4.56 * 36.5)/(17)
=9.79g of HCl
So the mass just need is
9.79g of HCl..![/quote]I xo much respect d workings u gave bro,buht am confused abt smfn....dou I got dat 40.58g of d HCl,bu I was contemplatin mayb I was wrong because d value has already exceed d initial value,.sir my Q now is dat weneva we found it as such,do we av 2 consider it 2 b enof 4 d reaction huh

...nd bsyds thanx for d disintegration

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 12:30am On Jan 07, 2014

[quote
author=Ai-bi-ke]I xo much respect d workings u gave bro,buht am
confused abt smfn....dou I got dat 40.58g of d HCl,bu I was contemplatin
mayb I was wrong because d value has already exceed d initial value,.sir
my Q now is dat weneva we found it as such,do we av 2 consider it 2 b
enof 4 d reaction huh

...nd bsyds thanx for d disintegration[/quote]Once the value has exceeded it,you don't have to it again...u use the one lesser....If you observe my working....u shall discover,i did not make use of it!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by DrLazDevitan: 12:32am On Jan 07, 2014

Dear Aysuccess,
The mail u sent through the NL is abortive.I cant get the mail...Sir!

Re: Online tutorial in Physics, Chemistry, Maths for Jamb and pume by labodinho: 1:00am On Jan 07, 2014

[quote author=Dr. Laz'Devitan][/quote]can i see your workings for the No.4 sir?

Tho you got all questions.

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