Class 12th

Aqueous solution of which of the following boron compounds will be strongly basic in nature?

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Class 12th Coordination Compounds

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Borax forms basic buffer solution in aqueous medium.

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Number of lone pair(s) of electrons on central atom and the shape of BrF₃ molecule respectively.

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Contributor-Level 10

Kindly consider the following figure

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Amongst baking soda, caustic soda and washing soda, carbonate anion is present

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Contributor-Level 10

Baking soda = NaHCO₃

Washing soda = Na₂CO₃. 10H₂O

Caustic soda = NaOH

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Given below are two statements:

Statement I : In CuSO₄.5H₂O, Cu-O bonds are present

Statement II : In CuSO₄.5H₂O, ligands coordinating with Cu(II) ion are O- and S- based ligands

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

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Contributor-Level 10

From the given structure of CuSO₄. 5H₂O Cu (II) ion and oxygen bonds are present but ligands coordinating with Cu (II) ion are not O and S both.

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6 months ago

Match List – I with List – II:

List – I (Ore) List – II (Composition)

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Contributor-Level 10

Ore Formula

Siderite FeCO₃

Malachite &n

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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) : The first ionization enthalpy for oxygen is lower than that of nitrogen

Reason (R) : The four electrons in 2p orbitals of oxygen experience more electron electorn repulsion.

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

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Class 12th Maths Determinants

Contributor-Level 10

1st ionization energy of N>O in oxygen atom, 2 electrons out of 4 electron of 2p orbital resulting in an increased electron repulsion.

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If $\vec{a} . \vec{b} = 1, \vec{b} . \vec{c} = 2 \vec{c} . \vec{a} = 3,$ then the value of $[\vec{a} * (\vec{b} * \vec{c}), \vec{b} * (\vec{c} * \vec{a}), \vec{c} * (\vec{b} * \vec{a})] i s :$

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$\vec{a} * (\vec{b} * \vec{c}) = 3 \vec{b} - \vec{c} = \vec{u}$

$\vec{b} * (\vec{c} * \vec{a}) = \vec{c} - 2 \vec{a} = \vec{v}$

$\vec{c} * (\vec{b} * \vec{a}) = 3 \vec{b} - 2 \vec{a} = \vec{w}$

$∴ \vec{u} + \vec{v} = \vec{w}$

so vectors

$\vec{u}, \vec{v} a n d \vec{w}$

are coplanar, hence their Scalar triple product will be zero.

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6 months ago

Class 12th Maths

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane in B(a, b, c), then

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Contributor-Level 10

Consider the equation of plane,

$P : (2 x + 3 y + z + 20) + λ (x - 3 y + 5 z - 8) = 0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So, $4 + 2 λ + 9 - 9 λ + 1 + 5 λ = 0$

0 $λ = 7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$(2, - \frac{1}{2}, 2)$

In plane P is (a, b, c) then

$\frac{a - 2}{1} = \frac{b + \frac{1}{2}}{- 2} = \frac{c - 2}{4}$

and $(\frac{a + 2}{2}) - 2 (\frac{b - \frac{1}{2}}{2}) + 4 (\frac{c + 2}{2}) = 4$

clearly

$a = \frac{4}{3}, b = \frac{5}{6} a n d c = - \frac{2}{3}$

So, a : b : c = 8 : 5 : 4

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6 months ago

Class 12th Maths

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

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Contributor-Level 10

$? l_{1} a n d l_{2}$ are perpendicular, so

$3 * 1 + (- 2) (\frac{α}{2}) + 0 * 2 = 0$

a = 3

Now angle between $l_{2} a n d l_{3}$ ,

$c o s θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt[]{1 + \frac{α^{2}}{4} + 4} . \sqrt[]{9 + 4 + 16}}$

$\Rightarrow c o s θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = c o s^{- 1} (\frac{4}{29}) = s e c^{- 1} (\frac{29}{4})$

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6 months ago

Class 12th Maths

Let the normal at the point P on the parabola y² = 6x pass through the point (5, 8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is

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