Class 11th

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$f (x) = s i n^{- 1} (\frac{3 x^{2} + x - 1}{{(x - 1)}^{2}}) + c o s^{- 1} (\frac{x - 1}{x + 1})$ $^{} \frac{^{}}{^{}}^{} \frac{}{}$

Domain of sin1 $(\frac{3 x^{2} + x - 1}{{(x - 1)}^{2}})$ $\frac{^{}}{^{}}$ is

$- 1 \leq \frac{3 s^{2} + x - 1}{{(x - 1)}^{2}} \leq 1$

$^{}^{}$ $- x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1$ and $3 x^{2} + x - 1 \leq x^{2} + 1 - 2 x$ $^{}^{}$

Domain of the function is $[\frac{1}{4}, \frac{1}{2}] \cup {0}$ .

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Class 11th Sets

If n is the number of solutions of the equation 2cos x $(4 s i n (\frac{π}{4} + x) s i n (\frac{π}{4} - x) - 1) = 1, x \in [0, π]$ and S is the sum of all these solution, then the order pair (n, S) is:

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Contributor-Level 10

$2 c o s x (4 s i n (\frac{π}{4} + x) s i n (\frac{π}{4} - x) - 1) = 1$

->2cos x (2 cos 2x – 1) = 1

$\Rightarrow c o s 3 x = \frac{1}{2}, F o r 0 \leq x \leq π, x = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}$

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Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Lithium salts are hydrated.

Reason (R) : Lithium has higher polarizing power than other alkali metal group members.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Lithium salts are extensively hydrated due to high hydration enthalpy of Li⁺

$L i^{+} > N a^{+} > K^{+} > R b^{+} > C s^{+}$ (order of polarizing power)

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The number of atoms in 8g of sodium is x * 10²³. The value of x is_______.

(Nearest integer)

[Given : N_A = 6.02 * 10²³ mol^-1, Atomic mass of Na = 23.0u]

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Contributor-Level 10

Moles of sodium = $\frac{8}{23} m o l$

Number of atoms = $\frac{8}{23} * 6.02 * 1 0^{23}$

$\approx 2 * 1 0^{23}$

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The spin-only magnetic moment value of species is________* 10^-2 BM. (Nearest integer)

[Given: $\sqrt[]{3} = 1.73]$

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$B_{2}^{+} \to σ 1 s^{2} s * 1 s^{2} σ 2 s^{2} σ * 2 s^{2} π 2 p^{1}$

Here, number of unpaired electrons, n = 1

Spin only moment ; $μ = \sqrt[]{n (n + 2)} B . M$

= 173 * 10^-2 B.M

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Which one of the following statements is incorrect?

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$H_{2} ?_{h v}^{h i g h ? ? T} 2 H$

Zn + 2HCl -> ZnCl₂ + H₂

$N a O H_{(a q)} + Z n ? N a_{2} Z n O_{2} + H_{2}$

$H_{2} \overset{2000 K}{?} 2 H$

It is nearly 0.081%.

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If 80g of copper sulphate CuSO₄ $\cdot$ 5H₂O is dissolved in deionized water to make 5L of solution. The concentration of the copper sulphate solution is x * 10^-3 mol L^-1. The value of x is________.

[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]

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Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

$C o n c e n t r a t i o n = \frac{m o l e s}{v o l u m e o f s o l u t i o n}$

$\begin{array}{l} = \frac{0.32}{5} = 0.064 M \\ = 64 * 1 0^{- 3} M \end{array}$

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The deposition of X and Y on ground surfaces is referred as wet and dry depositions, respectively, X and Y are:

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Class 11th Chemistry Chemical Bonding and Molecular Structure

Contributor-Level 10

Ammonium salt in rain drop resulting wet deposition

$N H_{4}^{+} ? S a l t + H_{2} O ? N H_{4} O H$

Oxides of N & S settle down on ground as dry deposition (SO₂).

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The molar solubility of Zn(OH)₂ in 0.1 M NaOH solution is x * 10^-18M. The value of x is______.

(Nearest integer)

(Given: The solubility product of Zn(OH)₂ is 2 * 10^-20)

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Contributor-Level 10

$Z n {(O H)}_{2} ? Z n^{2 +} + 2 O H^{-}$

S - -

NaOHNa⁺ + OH^-^-

0.1 M - -

$[Z n^{2 +}] = S$

Here ; 2 * 10^-20 = S (0.1)²

So; S = 2 * 10^-18 M

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