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Chemistry Class 12th

Match List – I with List – II:

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

...more

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Vishal Baghel

Contributor-Level 10

Ore Formula

Siderite FeCO₃

Malachite &n

...more

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a year ago

Chemistry Class 12th

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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Vishal Baghel

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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a year ago

Chemistry Class 11th

A 42.12% (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol is

[Given molar mass : Na = 23.0g mol^-¹; Cl = 35.5g mol^-¹]

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Vishal Baghel

Contributor-Level 10

Data is not sufficient but official answer is (D)

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a year ago

Chemistry Chemistry Chemical Bonding and Molecular Structure

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_P for the reaction is

[Given R = 0.082 L atm K^-¹ mol^-¹]

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Vishal Baghel

Contributor-Level 10

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

^{$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$}

^{$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$}

^{$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$}

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 00

2.5 PPP

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5 atm

$K_{P} = \frac{1.5 * 1.5}{1} = 2.25$

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a year ago

Chemistry Class 11th

Consider the species CH4, $N H_{4}^{+} a n d B H_{4}^{-} .$ $_{}^{}_{}^{}$ Choose the correct option with respect to the there species

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Vishal Baghel

Contributor-Level 10

Species =CH4 $_{}^{}$ $N H_{4}^{+}$ $B H_{4}^{?}$ $_{}^{}$

Electrons =101010

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Chemistry Class 11th

Which of the following is the correct plot for the probability density $ψ^{2} (r)$ $^{}$ as a function of distance 'r' of the electron from the nucleus for 2s orbital?

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Vishal Baghel

Contributor-Level 10

For 2S, number of radial moles = $n - l - 1 = 2 - 0 - 1 = 1$ and $^{}$ $Ψ^{2} (r)$ will always be positive

$∴ o p t i o n (B) i s c o r r e c t ?$ it has one radial node and it is positive.

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Chemistry Class 11th

Using the rules for significant figures, the correct answer for the expression $\frac{0.02858 * 0.112}{0.5702}$ $\frac{}{}$ will be

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Vishal Baghel

Contributor-Level 10

$\frac{0.02858 * 0.112}{0.5702} = 0.0561$

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a year ago

Chemistry Chemistry Chemical Bonding and Molecular Structure

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of $\frac{[C H_{3} C H_{2} C O O^{-}]}{[C H_{3} C H_{2} C O O H]}$ required to make buffer is_________.

Given : $K_{a} (C H_{3} C H_{2} C O O H) = 1.3 * 1 0^{- 5}$

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alok kumar singh

Contributor-Level 10

pH of acidic buffer is given by,

pH = pKa + log₁₀ $\frac{[S a l t]}{[a c i d]}$ $\frac{}{}$

or, $p = p k a + l o g_{10} \frac{[C H_{3} C H_{2} C O O^{-}]}{[C H_{3} C H_{2} C O O H]}$ $_{} \frac{_{}_{}^{}}{_{}_{}}$

or, $[H^{+}] = k_{a} \frac{[C H_{3} C H_{2} C O O H]}{[C H_{3} C H_{2} C O O^{-}]}$ $^{}_{} \frac{_{}_{}}{_{}_{}^{}}$

$\frac{[C H_{3} C H_{2} C O O^{-}]}{[C H_{3} C H_{2} C O O H]} = \frac{k_{a}}{[H^{+}]} = \frac{1.3 * 1 0^{- 5}}{1 0^{- 4}} = 0.13$

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a year ago

Chemistry Class 11th

In the structure of SF₄, the lone pair of electrons on S is in.

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alok kumar singh

Contributor-Level 10

Geometry of SF₄ is trigonal bipyramidal, in which there is one lonepair which occupy equatorial position as,

There are two lone pair – bond pair repulsions at 90°

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a year ago

Chemistry Structure of Atom

Consider the following statements:

(A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3,….

(B) The azimuthal quantum number $' l'$ for a given 'n' (principal quantum number) can have values as $' l' = 0, 1, 2, . . . n$

(C) Magnetic orbital quantum number $' m_{l}'$ for a particular $' l'$ (azimuthal quantum number) has $(2 l + 1)$ values.

(D) $\pm 1 / 2$ are two possible orientations of electron spin.

(E) For $l = 5,$ there will be a total of 9 orbitals

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alok kumar singh

Contributor-Level 10

$•$ Values of principal quantum number, n = 1, 2, 3, .

$•$ Values of azimuthal quantum number, $l = 0, 1, 2 . . . . . . . . . . ., n - 1$

$•$ Number of values of magnetic quantum number are $2 l + 1 .$

$•$ Values of spin quantum number are $\pm \frac{1}{2} .$

$•$ Number of orbitals for particular value of $l a r e 2 l + 1, s o f o r l = 5,$ no. of orbitals = 11.

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