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+

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.

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!

Dr. Laz'Devitan:


Dear sir...u are wrong......i saw so many errors that got u a bad landing!

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)....:...

great.david:
thanks for the correction sir.......

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.

Swagalord18:

these questions have no solution yet
I think the questions are wrong,

prove me wrong by providing the solution please

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

Swagalord18:

these questions have no solution yet
I think the questions are wrong,

prove me wrong by providing the solution please

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

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!

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!

labodinho:


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

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.

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

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

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!

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..!