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Why do alkanes not react with potassium permanganate?

Hydrocarbons with only single bonds are called alkanes. Alkanes are called saturated hydrocarbons because each carbon is bonded with as many... View Article

Which is more covalent PbCl2 or PbCl4

Pb is in +4 oxidation state in PbCl4 and in +2 oxidation state in PbCl2. The charge on the cation Pb is more in PbCl4 than in PbCl2. Therefore,... View Article

Which is more soluble in water NaCl or KCl

NaCl is more soluble in water than KCl. Solubility depends on the size of the cation as the anion is the same. The solubility of NaCl is 359 g/L... View Article

Which is more covalent NaCl or MgCl2

The covalent character is determined using Fajan’s rule. In the given compounds, the anion is the same and the cations are different. The size of... View Article

Does higher electronegativity mean stronger acid?

An atom will become more electronegative when the oxidation number increases. A higher value in electronegativity means a stronger acid.... View Article

If ? = 3 sin^-1 (6 / 11) and ? = 3 cos^-1 (4 / 9), where the inverse trigonometric functions take only the principal values, then the correct option (s) is (are)

1) cosβ > 0 2) sinβ < 0 3) cos(α + β) > 0 4) cosα < 0 Solution: (3) cos(α + β) > 0 ɑ = 3 sin-1 (6 / 11) > 3... View Article

The value of cot^-1 (9) + cosec^-1 (?41 / 4) is

1) π/2 2) π/4 3) π/3 4) π Solution: (2) π / 4 cot-1 (9) + cosec-1 (√41 / 4) = tan-1 (1 / 9) + tan-1 (4 / 5) = tan-1 (1 / 9) + (4... View Article

The value of tan {cos^-1 (- 2 / 7) – (π / 2)} is

1) 2/3√5 2) 2/3 3) 1/√5 4) 4/√5 Solution: (1) 2 / 3√5 tan {cos-1 (- 2 / 7) - (π / 2)} = tan {π - cos-1 (2 / 7) - (π / 2)} = tan {(π / 2) -... View Article

If sin^-1 (1 – x) – 2sin^-1 x = ? / 2, then x is equal to

1) 0, - (1/2) 2) 0, 1/2 3) 0 4) None of these Solution: (3) 0 sin-1 (1 - x) - 2sin-1 x = π / 2 sin-1 (1 - x) = π / 2 + 2 sin-1... View Article

The value of x, where x > 0 and tan {sec^-1 (1 / x)} = sin (tan^-1 2) is

1) √5 2) √5/3 3) 1 4) 2/3 Solution: (2) √5 / 3 Let {sec-1 (1 / x)} = θ sec θ = 1 / x cos θ = x tan {sec-1 (1 / x)} = tan θ = √1... View Article

If 2 tan^-1 (cos x) = tan^-1 (2 cosec x), then the value of x is

1) 3π/4 2) π/4 3) π/3 4) None of these Solution: (2) π/4 2 tan-1 (cos x) = tan-1 (2 cosec x) tan-1 (2cosx / [1 - cos2 x]) = tan-1... View Article

If sin^-1 x = ? / 5, for some x ? (-1, 1), then the value of cos^-1 x is

1) 3π/10 2) 5π/10 3) 7π/10 4) 9π/10 Solution: (1) 3π / 10 sin-1 x = π / 5 sin-1 x + cos-1 x = π / 2 (π / 5) + cos-1 x = π / 2... View Article

If sin^-1 x + sin^-1 y = ? / 2, then cos^-1 x + cos^-1 y =

1) π/2 2) π/4 3) π 4) 3π/4 Solution: (1) π / 2 sin-1 x + sin-1 y = π / 2 (π / 2) - cos-1 x + (π / 2) - cos-1 y = π / 2 cos-1 x +... View Article

If sin^-1 2a / [1 + a²] – cos^-1 [1 – b²] / [1 + b²] = tan^-1 2x / [1 + x²], then the value of x is

1) a 2) b 3) [a + b] / [1 - ab] 4) [a - b] / [1 + ab] Solution: (4) [a - b] / [1 + ab] sin-1 2a / [1 + a2] - cos-1 [1 - b2] / [1 +... View Article

If ?3 + i = (a + ib) (c + id), then tan^-1 (b / a) + tan^-1 (d / c) has the value

1) (π / 3) + 2nπ, n ∈ I 2) nπ + (π / 6), n ∈ I 3) nπ - (π / 3), n ∈ I 4) 2nπ - (π / 3), n ∈ I Solution: (2) nπ + (π / 6), n ∈ I √3... View Article

If sin^-1 (x / 5) + cosec^-1 (5 / 4) = ? / 2, then the value of x is

1) 1 2) 3 3) 4 4) 5 Solution: (2) 3 sin-1 (x / 5) + cosec-1 (5 / 4) = π / 2 sin-1 (x / 5) + sin-1 (4 / 5) = π / 2 sin-1 (x / 5)... View Article

tan^-1 x / (?a² – x²) =

1) 2 sin-1 (x / a) 2) sin-1 (2x / a) 3) sin-1 (x / a) 4) cos-1 (x / a) Solution: (3) sin-1 (x / a) Given, tan-1 x / (√a2 - x2) = θ... View Article

The solution of tan^-1 2? + tan^-1 3? = ? / 4 is

1) 1/√3 2) 1/3 3) 1/6 4) 1/√6 Solution: (3) ⅙ tan-1 2θ + tan-1 3θ = π / 4 tan-1 (2θ + 3θ) / (1 - 2θ × 3θ) = π / 4 5θ / [1 - 6θ2]... View Article

If sec^-1 (?1 + x²) + cosec^-1 (?1 + y²) / y + cot^-1 (1 / z) = 3?, then x + y + z =

1) xyz 2) 2xyz 3) xyz2 4) x2yz Solution: (1) xyz sec-1 (√1 + x2) + cosec-1 (√1 + y2) / y + cot-1 (1 / z) = 3π cos-1 1 / (√1 + x2)... View Article

The cos^-1 x = ?, (0 < x < 1) and sin^-1 (2x ?1 – x²) + sec^-1 1 / (2x² – 1) = 2? / 3, then tan^-1 (2x) =

1) π/6 2) π/4 3) π/3 4) π/2 Solution: (3) π / 3 cos-1 x = ɑ, (0 < x < 1) x = cos ɑ sin-1 (2x √1 - x2) + sec-1 1 / (2x2 -... View Article

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