Chemistry

Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface?

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Contributor-Level 10

During adsorption of a gas on solid surface, process is exothermic i.e. $Δ H = - v e (Δ H < 0)$ and entropy decreased, because gas particle resides on solid surface. Hence $Δ S < 0 .$

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5 months ago

Given below are two statements.

Statement I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charge on the nucleus as there is no strong hold on the electron by the nucleus.

Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.

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

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Contributor-Level 10

According to Bohr's model, velocity of electron increases according to atomic number (z).

$v_{e} \propto \frac{z}{n}$ (z = Atomic number and n = number of shell/principal quantum number)

Similarly, velocity of electron decreases according to number of shell (i.e. principal quantum number).

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5 months ago

83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is________K. (Nearest integer).

[Use: Molal Freezing point depression constant of water = 1.86 K kg mol^-1

Freezing point of water = 273K

Atomic masses: C : 12.0u, O : 16.0u, H : 1.0u]

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Contributor-Level 10

$Δ T_{f} = i * K_{f} * m$

{for ethylene glycol I = 1, due to non-ionisable}

$= 1 * 1.86 * \frac{83 * 1000}{62 * 625} K$

$= 3.984 K \approx 4 K$ (the nearest integer)

Hence, freezing point of solution = 273 K – 4K = 269 K.

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5 months ago

For water $Δ_{v a p} H = 41 k J m o l^{- 1}$ at 373K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is_________kJ mol^-1. [Use: R = 8.3 J mol^-1 K^-1]

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Contributor-Level 10

$Δ H_{v a p} = 41 K J m o l^{- 1}$

$Δ H = Δ E + Δ n R T$

$H_{2} O_{(l)} \to H_{2} O_{v a p}$

$Δ n = 1$

R = 8.3 JK^-1 mol^-1

T = 373 K

$Δ$ E (Change of internal energy) = $Δ H - Δ n R T$

$? Δ E = 41 - \frac{1 * 8.3 * 373}{1000} k J m o l^{- 1} = 41 - 3.09 k J m o l^{- 1} = 37.91 k J m o l^{- 1}$

Nearest integer = 38.

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5 months ago

The major product formed in the following reaction is:

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Kindly consider the following figure

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5 months ago

Given below are two statements.

Statement I: In the titration between strong acid and weak base methyl orange is suitable as an indicator.

Statement II: For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.

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

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Contributor-Level 10

Statement I is true because methyl orange indicator has pH range (3.2 to 4.2) which is suitable for buffer produced by strong acid and weak base.

Statement II is false because phenolphthalein has pH range (8.4 to 10.2) and only suitable for equivalence point above than 7, In this case pH increased by the addition of NaOH on acetic acid.

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5 months ago

Which one of the following methods is most suitable for preparing deionized water?

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Contributor-Level 10

De-ionised water means dissolved minerals free water which is prepared by synthetic resin method. Where cation exchanged by H⁺ and anion exchanged by OH^- of resin.

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5 months ago

Given below are two statements:

Statement I: Sphalerite is a sulphide ore of zinc and copper glance is a sulphide ore of copper.

Statement II: It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using 'depressants' in a froth flotation method.

Choose the most appropriate answer from the options given below:

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Contributor-Level 10

Sphalerite is the sulphide ore of zinc (ZnS) and copper glance Cu₂S is the sulphide ores of copper. Two sulphide ores can be separated by adjusting proportion of oil to water or by using depressant in froth flotation method.

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5 months ago

What are the products formed in sequence when excess of CO₂ is passed in slaked lime?

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Contributor-Level 10

$C a {(O H)}_{2} + C O_{2} \to C a C O_{3} + H_{2} O$ (1st step of reaction)

$C a C O_{3} + H_{2} O + C O_{2} \to C a {(H C O_{3})}_{2}$ (2nd step of reaction)

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5 months ago

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Photochemical smog causes cracking of rubber.

Reason (R): Presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in phtochemical smog makes it oxidizing.

Choose the most appropriate answer from the options given below:

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