Class 12th

Match List I with List II:

List I (molecule)		List II (hybridization; shape)
A.	XeO₃	I.	sp³d; linear
B.	XeF₂	II.	sp³; pyramidal
C.	XeOF₄	III.	sp³d³; distorted octahedral
D.	XeF₆	IV.	sp³d³; square pyramidal

Choose the correct answer from the options given below

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Contributor-Level 10

$X e O_{3} \Rightarrow s p^{3}, P y r a m i d a l$

XeF₂ sp³d₂, square pyramidal

XeF₆sp³d³, Distorted octahedral

XeOF₄ sp³d², square pyramidal

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5 months ago

SO₂Cl₂ on reaction with excess of water results into acidic mixture

SO₂Cl₂ + H₂O -> H₂SO₄ + 2HCl

16 moles of NaOH is required for the complete neutralization of the resultant acidic mixture. The number of moles of SO₂Cl₂ used is:

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Vishal Baghel

Contributor-Level 10

$S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l$

Let a moles of SO2Cl2 is taken

Then no. of moles of H2SO4 = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16

$\Rightarrow a = 4 m o l e s$

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The total number of Mn = O bonds in Mn₂O₇ is________.

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Number of Mn = O bonds = 6

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Given below are two statements.

Statement I: Iron (III) catalyst, acidified K₂Cr₂O₇ and neutral KMnO₄ have the ability to oxidize I^- to I₂ independently.

Statement II: Magnate ion is paramagnetic in nature and involves pp-pp bonding

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

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Contributor-Level 10

Neutral KMnO₄ oxidises I^- to and magnate ion has dpPp bonding

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Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula?

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Contributor-Level 10

Pyrophosphorous acid = H₄P₂O₅

Hypophosphorous acid = H₄P₂O₆

Phosphoric acid = H₃PO₄

Pyrophosphoric acid = H₄P₂O₇

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Class 12th Physics

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

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Contributor-Level 10

At any time $t$

$? = B A c o s ? \frac{π t}{5} (? ω = \frac{2 π}{T} = \frac{2 π}{10} = \frac{π}{5})$

$\frac{d ?}{d t} = - \frac{π}{5} B A s i n ? (\frac{π t}{5})$

$e = \frac{π}{5} B A s i n ? (\frac{π t}{5})$

e will be maximum when $\frac{π t}{5}$ is $\frac{π}{2}$ .

$t = 2.5 s e c$
e will be minimum when $\frac{π t}{5}$ is $π, 0$

$t = 0,5 s e c$

So, answer will be (1).

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Class 12th Maths Matrices

$2 s i n (\frac{π}{22}) s i n (\frac{3 π}{22}) s i n (\frac{5 π}{22}) s i n (\frac{7 π}{22}) s i n (\frac{9 π}{22}) i s e q u a l t o :$

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Contributor-Level 10

$2 s i n (\frac{π}{22}) s i n (\frac{3 π}{22}) s i n (\frac{5 π}{22}) s i n (\frac{7 π}{22}) s i n (\frac{9 π}{22})$

$= \frac{2 s i n \frac{32 π}{11}}{2^{5} s i n \frac{π}{11}} = \frac{1}{16}$

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5 months ago

Class 12th Maths

Let $\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} a n d l e t \hat{b}$ be a vector such that $\vec{a} * \vec{b} = 2 \hat{i} - \hat{k} a n d \vec{a}, \vec{b} = 3 .$ Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b} i s :$

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Contributor-Level 10

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

$\vec{a} * \vec{b} = 2 \hat{i} - \hat{k}$

$\vec{a} . \vec{b} = 3$

${| \vec{a} * \vec{b} |}^{2} + {| \vec{a} . \vec{b} |}^{2} = {| \vec{a} |}^{2} . {| \vec{b} |}^{2}$

$= \frac{\vec{b} . \vec{a} - {| \vec{b} |}^{2}}{| \vec{a} - \vec{b} |} = \frac{3 - \frac{7}{3}}{\sqrt[]{\frac{7}{3}}} = \frac{2}{\sqrt[]{21}}$

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Class 12th Physics

In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5 V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. The truth table of the given circuit will be:

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Vishal Baghel

Contributor-Level 10

When both A and B have logical value 'l' both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value '1'.

In all other case conduction will take place hence output will be zero value i.e. logical value 'o'.

So truth table is

A B Y

0 0

0 1 0 (AND gate)

...more

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Class 12th Maths

The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = Z a n d \frac{7 - x}{2} = y - 2 = z - 6$ is

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