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Class 11th Chemistry

Which pair oxides in acidic in nature?

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Vishal Baghel

Contributor-Level 10

Oxide Nature

CaO Basic

B₂O₃ Acidic

SiO₂ Acidic

BaO Basic

N₂O

...more

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Class 11th Chemical Bonding and Molecular Structure

Match List-I with List-II.

List-I List-II

(Molecule) &nb

...more

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Vishal Baghel

Contributor-Level 10

Species B.O

Ne₂ 0

N₂ 3

F₂ 1

O₂ &n

...more

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Class 11th Chemistry Chemical Bonding and Molecular Structure

Which of the following are isostructural pairs?

A. $S O_{4}^{2 -} a n d C r O_{4}^{2 -}$

B. SiCl₄ and TiCl₄

C. NH₃ and $N O_{3}^{-}$

D. BCl₃ and BrCl₃

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alok kumar singh

Contributor-Level 10

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Class 11th Chemistry Hydrocarbon

Which of the following reagent is used for the following reaction?

$C H_{3} C H_{2} C H_{3} \overset{?}{\to} C H_{3} C H_{2} C H O$

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3 months ago

Class 11th Hydrocarbons

In $\overset{1}{C} H_{2} = \overset{2}{C} = \overset{3}{C} H - \overset{4}{C} H_{3}$ molecule, the hybridization of carbon 1,2,3 and 4 respectively are:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Class 11th Chemistry

(A) HOCl + H₂O_2-> H₃O⁺ + Cl^- + O₂

(B) $I_{2} + H_{2} O_{2} + 2 O H^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}$

Choose the correct option.

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alok kumar singh

Contributor-Level 10

In both reactions H₂O₂ act as reducing agent

In reaction (A) HOCl – O.A

H₂O₂ – R.A

In reaction (B) I₂ – O.A

H₂O₂ – R.A

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Class 11th Chemistry

Which of the following forms of hydrogen emits low energy β - particles?

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Vishal Baghel

Contributor-Level 10

Tritium is radioactive and it decays into He³ during emission of β-radiation

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Class 11th Maths

In an increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of 4^th, 6^th and 8^th terms is equal to

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Payal Gupta

Contributor-Level 10

Let a, ar, ar², . an increasing G.P then r > 1 & a > 0

given : ar + ar5 = $\frac{25}{2}$ .(i)

and $a r^{2} a r^{4} = 25 \Rightarrow a^{2} r^{6} = 25 \Rightarrow {(a r^{3})}^{2} = 25$

$\Rightarrow a r^{3} = 5 . . . . . . . (i i)$

From (i) and (ii), $\frac{a r (1 + r^{4})}{a r^{3}} = \frac{25}{2 * 5}$

$\Rightarrow 2 + 2 r^{4} = 5 r^{2} \Rightarrow 2 r^{4} - 5 r^{2} + 2 = 0$

$\Rightarrow (2 r^{2} - 1) (r^{2} - 2) = 0$

$\Rightarrow r^{2} = \frac{1}{2}, 2 ∴ r^{2} = 2 \Rightarrow r = \sqrt[]{2}$

$∴ t_{4} + t_{6} + t_{8} = 5 + 10 + 20 = 35$

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Class 11th Maths

The maximum value of term independent of 't' in the expansion of ${(t x^{\frac{1}{5}} + \frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{10}$ where $x \in (0, 1) i s :$

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Payal Gupta

Contributor-Level 10

$T_{r + 1} =^{10} C_{r} {(t x^{\frac{1}{5}})}^{10 - r} {(\frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{r}$

According to question, 10 – 2r = 0 ⇒ r = 5

$∴ T_{6} =^{10} C_{5} * {(1 - x)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x {(1 - x)}^{\frac{1}{2}}$ is maximum.

$f' (x) = {(1 - x)}^{\frac{1}{2}} - \frac{x}{2 \sqrt[]{1 - x}} = \frac{2 (1 - x) - x}{2 \sqrt[]{1 - x}}$

For maximum, $f' (x) = 0 \Rightarrow x = \frac{2}{3}$

$∴ T_{6} =^{10} C_{5} \frac{2}{3} . {(\frac{1}{3})}^{\frac{1}{2}} = \frac{2 (10!)}{3 \sqrt[]{3} {(5!)}^{2}}$

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Class 11th Maths

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + . . . . .$ is equal to

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Payal Gupta

Contributor-Level 10

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵+ .

$\underline{x S = x + 2 x^{2} + 7 x^{3} + 12 x^{4} + 17 x^{5} + . . . . . . . . .}$

Subtracting,

$(1 - x) S = 1 + x + 5 x^{2} + 5 x^{3} + 5 x^{4} + 5 x^{5} + . . . . . .$

= $1 - 4 x + 5 x + 5 x^{2} + 5 x^{3} + . . . . . . . . .$

$= \frac{1 - x - 4 x + 4 x^{2} + 5 x}{1 - x} = \frac{4 x^{2} + 1}{1 - x}$

$S = \frac{4 x^{2} + 1}{{(1 - x)}^{2}} p u t x = \frac{1}{3} w e g e t s = \frac{\frac{4}{9} + 1}{\frac{4}{9}} = \frac{13}{4}$

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