Class 11th

The number of sigma bonds in

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Raj Pandey

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Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]

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The number of real solutions of the equation, x² - |x| - 12 = 0 is:

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Contributor-Level 10

x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.

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Let the equation of the pair of lines, y = px and y = qx, can be written as (y - px)(y - qx) = 0. Then the equation of the pair of the angle bisectors of the lines x² - 4xy – 5y² = 0 is:

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Contributor-Level 10

x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0

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The sum of all those terms which are rational numbers in the expansion of (2¹/³ + 3¹/?)¹² is:

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Contributor-Level 10

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Given below are two statements:
Statement I: None of the alkaline earth metal hydroxides dissolves in alkali.
Statement II: Solubility of alkaline earth metal hydroxides in water increases down the group.
In the light of the above statements, choose the most appropriate answer from the options given below:

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Raj Pandey

Contributor-Level 9

Solubility of 2 group hydroxide increases down the group.

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Class 11th Thermodynamics

For water at 100°C and 1 bar, ΔᵥₐₚH - ΔᵥₐₚU = _______ * 10² J mol⁻¹ (Round off to the Nearest integer) [Use: R = 8.31 J mol⁻¹ K⁻¹] [Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) can be treated as an ideal gas]

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Contributor-Level 10

H? O (l) → H? O (g)
ΔH° = ΔU° + ΔngRT
ΔH° - ΔU° = ΔngRT
= 1 * 8.31 * 373
= 3099.63 J/mol
= 30.9963 * 10² J/mol
≈ 31 * 10² J/mol

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If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

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Contributor-Level 10

Equation of tangent of P (2cosθ, sinθ) is
(cosθ)x + (2sinθ)y = 4
Solving equation of tangent with equation of tangents at major axis ends, i.e. x = -2 and x = 2
For point 'B' (at x=-2):
-2cosθ + 2sinθ y = 4 ⇒ y = (2+cosθ)/sinθ
B (-2, (2+cosθ)/sinθ)
For point 'C' (at x=2):
2cosθ + 2sinθ y = 4 ⇒ y = (2-cosθ)/sinθ
C (2, (2-cosθ)/sinθ)
Now BC is the diameter of circle
Equation of circle: (x+2) (x-2) + (y - (2+cosθ)/sinθ) (y - (2-cosθ)/sinθ) = 0
x²-4 + y² - (4/sinθ)y + (4-cos²θ)/sin²θ = 0
Check if (√3, 0) satisfies this:
(√3)²-4 + 0 - 0 + (4-cos²θ)/sin²θ = -1 + (3+sin²θ)/sin²θ = -1 + 3/sin²θ + 1 = 3/sin²

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Class 11th Solutions

The density of NaOH solution is 1.2 gm cm⁻³. The molality of this solution is _______ m. (Round off to the Nearest integer) [Use: Atomic masses: Na 23.0u O: 16.0 u H:1.0 u Density of H₂O: 1.0g cm⁻³]

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Contributor-Level 10

Let volume of solution = x ml
So mass of solution = 1.2x
And mass of water = x gm
Mass of solute = 0.2x
Molality = (W_solute * 1000) / (M_solute * W_solvent) = (0.2x * 1000) / (40 * x) = 5 m

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The difference between bond orders of CO and NO⁺ is x/2 where x is _______. (Round off to the Nearest integer)

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Contributor-Level 10

B.O. of CO = 3
B.O. of NO? = 3
Both are isoelectronic
So difference = 0
∴ x = 0

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PCl₅ ⇌ PCl₃ + Cl₂ Kc = 1.844. 3.0 moles of PCl₅ is introduced in a 1L closed reaction vessel at 380K. The number of moles of PCl₅ at equilibrium is _______ * 10⁻³. (Round off to the Nearest integer)

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