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Surface Chemistry Class 12th

Given below are the critical temperature of some of the gases:

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Surface Chemistry Class 12th

The separation of two coloured substances was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 3.25cm, 2.08cmand 1.05cm, respectively. The ratio of R_f values of A to B is ________.

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Payal Gupta

Contributor-Level 10

$R_{f}^{0} = distance travelled by the solute$ $/$ $/ distance travelled by the solvent$

${(R_{f})}_{A} = \frac{2.08}{3.25} {(R_{f})}_{B} = \frac{1.05}{3.25}$

$\frac{{(R_{f})}_{A}}{{(R_{f})}_{B}} = \frac{2}{1}$

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2 months ago

Surface Chemistry Class 12th

Incorrect statement for Tyndall effect is:

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Payal Gupta

Contributor-Level 10

$A l^{+ 3}$ → number of electrons = 10

Br → number of electrons = 36

$B e^{+ 2}$ → number of electrons = 2

$C l^{-}$ → number of electrons = 18

$L i^{+}$ → number of electrons = 2

$S^{- 2}$ → number of electrons = 18

K⁺ → number of electrons = 18

O^-2 → number of electrons = 10

Mg⁺² → number of electrons = 10

O^-2 & Mg⁺² are isoelectronic with Al⁺³

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Surface Chemistry Class 12th

When 200mL of 0.2 M acetic acid is shaken with 0.6g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1M. The mass of acetic acid adsorbed per gram of carbon is ______________ g.

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Payal Gupta

Contributor-Level 10

Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$= (0.2 * 200) * 1 0^{- 3} - (0.1 * 200) * 1 0^{- 3}$

= 20 * 10^-3 moles

Mass of CH₃COOH absorbed = 20 * 10^-3 * 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6} = 2 g$ of charcoal.

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Surface Chemistry Class 12th

Given below are two statements :

Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing.

Statement II : For stabilization of an emulsion, excess of electrolyte is added. In the light of the above statements, choose the most appropriate answer from the options given below :

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Vishal Baghel

Contributor-Level 10

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

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Surface Chemistry Class 12th

Hydrogen has three isotopes : protium (¹H), deuterium (²H or D) and tritium (³H or T). They have nearly same chemical properties but different physical properties. They differ in

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alok kumar singh

Contributor-Level 10

Isotopes are the atom of same element with different atomic mass.

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Surface 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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3 months ago

Surface Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Which one of the following is an example of disproportionation reaction?

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Vishal Baghel

Contributor-Level 10

Disproportionation reaction is

$\overset{+ 6}{M g} O_{4}^{- -} + 4 H^{+} \to 2 \overset{+ 7}{M n} O_{4}^{-} + \overset{+ 4}{M n} O_{2} + 2 H_{2} O$

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3 months ago

Surface Chemistry Physics NCERT Exemplar Solutions Class 12th Chapter Five

Consider the ions /molecule $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2 -}$

For increasing bond order the correct option is

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Payal Gupta

Contributor-Level 10

$B o n d o r d e r = \frac{e^{-} o f B o n d i n g M O - e^{-} o f A B M O}{2}$

Hence Bond order for $O_{2}^{+} = \frac{10 - 5}{2} = 2.5$

$O_{2} = \frac{10 - 6}{2} = 2.0$

$O_{2}^{-} = \frac{10 - 7}{2} = 1.5$

$O_{2}^{- -} = \frac{10 - 8}{2} = 1.0$

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3 months ago

Surface Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

Which of the following is a correct statements?

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alok kumar singh

Contributor-Level 10

Due to equal and similar charge particle will repel each other, hence will never precipitate

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