Class 12th

If vertex of a parabola is (2, -1) and the equation of its directrix is 4x – 3y = 21, then the length of its latus rectum is:

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Class 12th Chemistry Chemical Kinetics

Contributor-Level 10

$a = | \frac{4.2 - 3 (- 1) - 21}{5} |$

$= | \frac{8 + 3 - 21}{5} | = 2$

$∴ L = 4 a = 8$

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8 months ago

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 – 1000 K. The data has been analysed by plotting ln k vs $\frac{1 0^{3}}{T}$ graph. The value of activation energy for the reaction is ___________ kJ mol^-¹.

(Nearest integer)

(Given : R = 8.31 J K^-¹ mol^-¹)

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Contributor-Level 10

$k = A . e^{- (\frac{E a}{R T})}$

$l n k = l n A - \frac{E a}{R T}$

$\Rightarrow l n k = l n A + (- \frac{E a}{1000 R}) . \frac{1000}{T}$

Slope = $\frac{- E a}{1000 R} = - 18.5$

=> Ea = 18.5 * 1000 * 8.31 = 153.735 * 10³ J = 154 KJ

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8 months ago

Class 12th Electrochemistry

The cell potential for the following cell

$P t | H_{2} (g) | H^{+} (a q) | | C u^{2 +} (0.01 M) | C u (s)$

Is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)

(Given : $E_{C u^{2 +} / C u}^{0} = 0.34 V a n d \frac{2.303 R T}{F} = 0.06 V$ )

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Contributor-Level 10

Anode, H₂ (g) -> 2H⁺ + 2e^-

Cathode, Cu²⁺ + 2e -> Cu (s)

Net cell reaction H₂ + Cu²⁺ -> 2H⁺ Cu (s)

$E_{c e l l}^{0} = E_{C u^{2 +} / C u}^{0} - E_{H^{+} / H_{2}}^{0}$

= 0.34 V – 0

= 0.34 V

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.06}{2} l o g \frac{{[H^{+}]}^{2}}{[C u^{2 +}]}$

$\begin{array}{l} \Rightarrow 0.576 = 0.34 - \frac{0.06}{2} l o g \frac{{[H^{+}]}^{2}}{0.01} \\ \Rightarrow - l o g [H^{+}] = 4.93 \\ \Rightarrow p H o f s o l u t i o n = 4.93 ~ 5 \end{array}$

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8 months ago

Let $\vec{a} = α \hat{i} + 2 \hat{j} - \hat{k} a n d \vec{b} = - 2 \hat{i} + α \hat{j} + \hat{k}, w h e r e α \in R .$ If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a} a n d \vec{b} i s \sqrt[]{15 (α^{2} + 4)},$ then the value of $2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2}$ is equal to:

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Contributor-Level 10

$\vec{a} = α \hat{i} + 2 \hat{j} - \hat{k} & \vec{b} = - 2 \hat{i} + α \hat{j} + \hat{k}$

$| \vec{a} * \vec{b} | = \sqrt[]{15 (a^{2} + 4)}$

$2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2} = ?$

$\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ$

$c o s θ = \frac{- 1}{(a^{2} + 5)}$ $∴ | s i n θ | = \frac{\sqrt[]{{(a^{2} + 5)}^{2} - 1}}{(a^{2} + 5)}$

$| \vec{a} * \vec{b} | = | \overset{?}{a} | * | \vec{b} | * | s i n θ |$

$= (a^{5} + 5) * \frac{\sqrt[]{{(a^{2} + 5)}^{2} - 1}}{(a^{2} + 5)} = \sqrt[]{15 (a^{2} + 4)}$

${(a^{2} + 5)}^{2} - 1 = 15 (a^{2} + 4)$

$\Rightarrow (α = \pm 3)$

$2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2}$

= $2 (a^{2} + 5) - (a^{2} + 5)$

$(a^{2} + 5) = 14$

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8 months ago

Class 12th Solutions

The osmotic pressure of blood is 7.47 bar at 300K. To inject glucose to a patient intravenously, it has to isotonic with blood. The concentration of glucose solution in gL^-¹ is __________.

(Molar mass of glucose = 180 g mol^-¹ R = 0.083 L bar K^-¹ mol^-¹) Nearest integer)

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Contributor-Level 10

For isotonic solution

$?_{i n j e c t o r} = ?_{B l o o d}$

$? C * 0.082 * 300 = 7.47$

$? C = 0.3 ? ? m o l e / l$

$= 0.3 * 180 g m / l = 54 g m / l$

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8 months ago

Class 12th Chemistry

Transition metal complex with highest value of crystal field splitting $(Δ_{0})$ will be

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Contributor-Level 10

The value of crystal field spilitting energy $_{}$ $(Δ_{0})$ increases down the group. 5d series member will have more $_{}$ $Δ_{0}$ compared to 3d series or 4d series member.

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8 months ago

Class 12th Chemistry

Atoms of element X from hcp lattice and those of element Y occupy $\frac{}{}$ of its tetrahedral voids. The percentage of element X in the lattice is ________. (Nearest integer)

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Contributor-Level 10

Unit cell = Hexagonal close packing (hcp)

No. of X in one unit cell = 6

No. of tetrahedral voids in one unit cell = 12

No. of Y in one unit cell = $\frac{2}{3} * 12 = 8$ $\frac{}{}$

Formula => X₆ Y₈ => X₃ Y₄

% of X in unit cell $= \frac{3}{7} * 100 % = 43 %$ $\frac{}{}$

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8 months ago

Let a > 0, b > 0. Let a and respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ Let e' and $l'$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^{2} = \frac{11}{14} l a n d {(e')}^{2} = \frac{11}{8} l',$ then the value of 77a + 44b is equal to

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Contributor-Level 10

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$e = \sqrt[]{1 + \frac{b^{2}}{a^{2}}}$ $e' = \sqrt[]{1 + \frac{a^{2}}{b^{2}}}$

$l = \frac{2 b^{2}}{a}$ $l' = \frac{2 a^{2}}{b}$

$(1 + \frac{b^{2}}{a^{2}}) = \frac{11}{14} * \frac{2 b^{2}}{a}$ $(1 + \frac{a^{2}}{b^{2}}) = \frac{11}{8} * \frac{2 a^{2}}{b}$

$7 * {{(\frac{7 b}{4})}^{2} + b^{2}} = 11 * b^{2} * \frac{7 b}{4} \Rightarrow a * 77 = 65$

65b² = 44b³

65 = b * 44

$∴ 77 a + 44 b = 65 * 2 = 130$

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8 months ago

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : -4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the nation 1 :3, then the area of the triangle is equal to:

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Contributor-Level 10

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Class 12th Chemistry

During the qualitative analysis of salt with cation y²⁺, addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation (y²⁺) present represent respectively are:

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