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L.H.S. = \(\frac{tanx + sec x-1}{tan x-sec x+1}\) = \(\frac{2tan x}{2sec x-2}\) = \(\frac{tan x}{sec x-1}\)

 = \(\frac{sinx}{1-cos x}\) = \(\cfrac{2sin\frac x2cos\frac x2}{2sin^2\frac x2}\) = cot x/2

R.H.S. = \(\frac{1+sin x}{cos x}\) = \(\cfrac{(sin \frac x2+cos\frac x2)^2}{(cos\frac x2+sin \frac x2)(cos\frac x2-sin\frac x2)}\)

 = \(\frac{sin\frac x2+cos\frac x2}{cos\frac x2-sin\frac x2}\) =  \(\frac{2cos\frac x2}{2sin\frac x2}=\) cot x/2

Hence, L.H.S. = R.H.S.

⇒ \(\frac{tanx + sec x-1}{tan x-sec x+1}\) = \(\frac{1+sinx}{cos x}\) Hence Proved

5678946+868685 = 6547631

L.H.S. = \(\frac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}\) = \(\frac{(1+cos\theta+sin\theta)(1+cos\theta-sin\theta)}{(1+cos\theta+sin\theta)-(1+cos\theta-sin\theta)}\)

 = \(\frac{2(1+cos\theta)}{2sin\theta}\) 

 = \(\frac{(1+cos\theta)(1-cos\theta)}{sin\theta(1-cos\theta)}\) 

= \(\frac{1-cos^2\theta}{sin\theta(1-cos\theta)}\) = \(\frac{sin^2\theta}{sin\theta(1-cos\theta)}\)

= \(\frac{sin\theta}{1-cos\theta}\) = \(\cfrac{2sin\frac{\theta}2cos\frac{\theta}2}{2sin^2\frac{\theta}2}\) 

 L.H.S. = cot θ/2

R.H.S. = \(\frac{1+sin\theta}{cos\theta}\)

\(=\frac{sin^2\theta/2+cos^2\theta/2+2sin\theta/2cos\theta/2}{cos^2\theta/2-sin^2\theta/2}\)

\(=\frac{(cos\theta/2+sin\theta/2)^2}{(cos\theta/2+sin\theta/2)(cos\theta/2-sin\theta/2)}\) 

\(=\frac{cos\theta/2+sin\theta/2}{cos\theta/2 - sin\theta/2}\)

\(= \frac{(cos\theta/2+sin\theta/2)+(cos\theta/2-sin\theta/2)}{(cos\theta/2+sin\theta/2)-(cos\theta/2-sin\theta/2)}\)

\(=\frac{2cos\theta/2}{2sin\theta/2}=cot\theta/2\)

Hence, L.H.S = R.H.S.

∴ \(\frac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}\) = \(\frac{1+sin\theta}{cos\theta}\)

Hence Proved

\(tan \frac\theta4 = \cfrac{\frac\mu2sin\theta}{\frac\mu2 + \frac\mu2cos\theta} = \frac{sin\theta}{1+ cos\theta}\)

\(= \cfrac{2sin\frac\theta2cos\frac\theta2}{2cos^2\frac\theta2}\)

\(= tan \left(\frac\theta2\right)\)

\(\therefore \frac\theta4 = \frac\theta2 \)

⇒ \(\theta = 0\)

tan θ/4 = tan θ/2

⇒ θ/4 = nπ + θ/2

⇒ θ/4 - θ/2 = nπ

⇒ -θ/4 = nπ

⇒ θ = -4nπ, n ∈ \(\mathbb{Z}\)

√61 units will be the distance between the points.

Steel is a mixture.

1. Number of moles in 1 g water = \(\frac{1}{8}\)

∴ No. of molecules in 1 g water

= \(\frac{1\times6.022\times10^{23}}{18}\) = 3.35 x 10²²

2. Volume of 0.2 mol SO₂ at STP = 0.2 × 22.4 litre 

= 4.48 litre

1 µg = 10^-6g

(a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively. 

First Oxide : 

Mass of oxygen = 100 – 78.7 = 21.3 g 

No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{78.7}{21.3}\) = 3.7g

Second oxide: 

Mass of oxygen = 100 – 64.5 = 35.5 g 

No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}\) = 1.9g

The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

(b) 1/12^th the mass C¹² atom – amu 

1 g of hydrogen atom – gram atom 

22.4 L O₂ at NTP – molar volume

180 g of glucose – gram mole 

6.022 × 10²³ particles – 1 mole

The number of significant figures in 0.00503060 is 6.

1. Law of multiple proportions. 

When two elements combine to form more than one compound the different mass of one of the elements which combine with the fixed mass of the other element bear a simple ratio.

2. 

In NO and NO₂ , the masses of oxygen combining with a fixed mass (14 g) of nitrogen are in the ratio, 16:32 = 1:2. Similarly, in N₂O and N₂O₃ , the masses of oxygen combining with a fixed mass (28 g) of nitrogen are in the ratio, 16:48 = 1:3. These are simple whole number ratios. Hence, the law of multiple proportions is verified.

Correct answer is (c) babies

(b) Law of conservation of mass

Correct answer is (b) depression

The compound will be N₂O₄

(d) 44 g carbon dioxide

Correct answer is (a) wild

0.1254000000000

(a) A song is being sung by Mina.

(b) By whom was the Ramayan written?

(c) The thief was caught.

Gram atomic mass of He = 4 g. 

Thus, number of atom sin 4g (1 mol) He = 6.02 × 10²³ 

So number of atoms in 48 g of He = \(\frac{48}{4}\) x 6.02 × 10²³

=12 × 6.02 × 10²³ 

= 7.224 × 10²⁴

The ratio of gram atoms of Au and Cu in 22ct gold is 7 : 2

(1) Take dil.soap solution in a test tube. 

(2) Add grease drops to test tube. 

(3) Just shake the test tube. 

(4) Micelles will be formed in the test tube.

\(\underset{C(1 -\alpha)}{CH_3COOH} \rightleftharpoons \underset{C\alpha}{CH_3COO^\ominus} + \underset{C\alpha}{H^+}\)

\(K_a = \frac{C^2\alpha^2}{C(1 - \alpha)}\)

\(K_a = \frac{C\alpha^2}{1 - \alpha}\)

∵ Acetic acid is very weak acid, so we can write 1 - α ≈ 1

∴ K_a = cα²

\(\alpha = \sqrt{\frac{K_a}C}\)

As we can see in the above relation degree of dissociation of Acetic acid depends on the concentration of acid and its dissociation constant.

As we know dissociation constant only depends on the temperature therefore, degree of dissociation also affected by temperature.

At constant temperature, when concentration of acid decreases the degree of dissociation of Acetic acid increases.

Hence, temperature and dilution of Acetic acid affect the degree of dissociation.

X₂^-  is having 15 electrons = no. of protons 

Hence X=Nitrogen

Z=7

Number of neutrons = 8

might be answer is incorrect bcoz u hv not written ques in a right manner

What is Kohlrausch’s Law?

Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.

The molar conductivity of a solution at a given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length. The molar conductivity of a solution increases with the decrease in concentration. This increase in molar conductivity is because of the increase in the total volume containing one mole of the electrolyte. When the concentration of the electrolyte approaches zero, the molar conductivity is known as limiting molar conductivity, Ëm°.

Coordination isomers are two or more coordination compounds in which the composition within the coordination sphere (i.e., the metal atom plus the ligands that are bonded to it) is different (i.e., the connectivity between atoms is different).

The process of converting a freshly prepared precipitate into colloidal form by the addition of a suitable electrolyte in small amount is called peptization.

Peptization involves the adsorption of suitable ions from the electrolyte by the particles of precipitate

Fe(OH)₃ + Fe³⁺ → [Fe(OH)₃]Fe³⁺

(a) Trans-4-bromopent-2-ene

(b) Cis-1-bromo-2-methylbut-2-ene 

There is no vacant d-orbital in the outermost orbit of Nitrogen. Thus nitrogen show valency only three. There are valent d-orbitals in the outer most orbit of phosphorus and hence it shows variable covalence 3 and 5 in ground state and excited state respectively. Hence nitrogen forms only NCl₃ but phosphorus forms PCl₃ and PCl₅ both. 

Any chemical substance which is formed in the tissues of endocrine glands are carried by the blood to other parts of the body for its specific actions is termed as hormone. Plant growth promoting hormone is auxin :

–Movement of sensitive leaves is called as Nastism. It is unidirectional movement of leaves to stimulus.e.g. touch-me-not plant leaves. 

–Movement of shoot towards light is called as tropism. It is directional movement of shoot towards light.

(a)  A is non- metal

(b)  C is less reactive then A

(c)  C will be smaller then B because nuclear charge will increases. 

Oxidation

1. Addition of oxygen

2. C + O₂ → CO₂

Reduction

1.   Removal of oxygen

2.  ZnO + C +  → Zn + CO

Magnesium ribbon is a very reactive metal when stored. It reacts with oxygen to form a layer of magnesium oxide (MgO) on its surface. This layer of magnesium oxide being a stable compound prevents further reaction of magnesium with oxygen.