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Differences between O.N. and Valency

O.N.	Valency
It is the charge appeared on the atom. eg: O.N. of C in HCHO is 0.	It is the no. of electrons lost, gained or shared by an atom of an element. eg: Valency of C in HCHO is 4.
It may be zero, positive, negative or even in fraction. eg: O.N. of N in (hydrozoic acid) N3H is -1/3 and in NH3is -3.	It is a whole number with positive or negative sign. eg: Valency of N in N3H and NH3is 3.
It may be different for the same element in the different compounds. eg: O.N. of C in CH4, CH3Cl,...

Carbon combines with hydrogen to form three binary compounds X, Y and Z. The percentage of hydrogen in X, Y and Z is 25%, 14.3% and 7.7% respectively, which law of chemical combination is illustrat...

Solution:
Given,
In the compound X,
% of hydrogen = 25% & % of carbon = (100 – 25)% = 75% 25 parts by weight of hydrogen combine with 75 parts by weight of carbon.
1 part by weight of hydrogen combine withparts by weight of carbon
= 3 parts by weight of carbon

In the compound Y,
% of hydrogen = 14.3% & % of carbon = (100 – 14.3)% = 85.7% 14.3 parts by weight of hydrogen combine with 85.7 parts by weight of carbon
1 part by weight of hydrogen combine withparts by weight of carbon
=5.99 parts by...

Phosphorous and oxygen combined to each other to give PO and PO. Which law of stoichiometry is illustrated. Also state the law.

Here,
In P₂O₃, 62gm of Phosphorous combined with 48gm of oxygen.
In P₂O₅, 62gm of Phosphorous combined with 80gm of oxygen.

Ratio of mass of oxygen in P₂O₃and P₂O₅= 48:80 = 3:5 which is simple whole number ratio.

In this reaction, the ratio of mass of oxygen in PO and PO combined with fixed mass of phosphorous bear simple whole number ratio.

This illustrates, the law of multiple proportion of stoichiometry which states that,"Whentwo elements combined to each other to give two or more productsthen...

In an experiment 0.936 g of zinc was converted into the oxide and the weight of oxide formed was1.165 g. In a second experiment 1.236 g of the metal was converted into the oxide and the weight ofox...

Ans:Solution:
Given,
Experiment 1,

Wt. of zinc = 0.936 g
Wt. of oxide = 1.165 g
Wt. of oxygen = (1.165 – 0.936) g = 0.229 g

The ratio of weight of oxygen is to zinc in zinc oxide is

Experiment 2,

Wt. of zinc = 1.236 g
Wt. of oxide = 1.538 g
Wt. of oxygen = (1.538 – 1.236) g = 0.302 g

The ratio of weight of oxygen is to zinc in this oxide is

The ratio of weight of oxygen to weight of zinc in both the oxides is the same. Hence, given data
illustrate the law of definite proportions.

Alternative Method

In...

12g of Carbon combines with 32g of oxygen to give 44g of carbon dioxide. Which law of stoichiometry is illustrated? And also state the law.

Ans:
In this reaction, the total mass of reactants is equal to the total mass of the product.

This illustrates the law of conservation of mass which states, "Under similar conditions of temperature and pressure, the total mass of reactants in any chemical reaction is equal to the total mass of product."

The Cosine Law

Proof:
We consider a triangle ABC then we can have three possible figures where angle C is acute in figure i, right angle in figure ii and obtuse in figure iii.

From A draw AD perpendicular to BC ( Produce BC if necessary )

In ΔABC, AB²= AD²+ BD²
c²=AD²+ BD² Ⅰ

In figure (i),
AD = bSinC

Also,
DC = ACCosC = bCosC
BD = BC - DC = a-bCosC

Now from equation Ⅰ,
c²= AD²+ BD²(becomes)
c²= (bSinC)²+ (a-bCosC)²
c²= b²Sin²C+ a²- 2abCosC + b²CosC²
c²= b²+ a²- 2abCosC
2abCosC = b²+ a²- c²

Similarly in figure ii,

AB²= BC^2...

The Sine Law in terms of Circum-Radius R.Statement: In any ΔABC, where R is the Circum-Radius of ΔABC.

Proof:
Consider the circum-circle of ΔABC with center O and radius R. Then we can have three possible figures.

Where the angle A is acute in fig i, or right angled in fig ii or obtuse in fig iii.
Let O be the center of the circum-circle of ΔABC and R be the circum-radius. Join BO and produce it to meet the circle at D.

Now in figure i,
BAC = BDC = A, BCD = 90º (angle at the semicircle).

= Sin(BDC)
= SinA
2R

In fig iii,
1 = Sin90º = SinA
SinA ( Since BC = BD )
SinA
2R

In fig iii,SinA
SinA
2R

Therefore in each...