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NEET chemistry

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Question
CBSEENCH11008422
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Calculate the pOH of a solution at 250C that contains 1 x 10-10 M of hydronium ions.

  • 7

  • 4

  • 9

  • 1

Solution

B.

4

[H3O+] = [H+] = 10-10

pH + pOH = [14]
pH = - log [ H+]
pH = - log [10-10]
pH = 10
⇒ pOH + 10 = 14
⇒ pOH = 14-10 = 4

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Question
CBSEENCH11008424
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A Weak acid, HA has a Ka of 1.00 x 10-5. If 0.100 mole of this percentage of acid dissociated at equilibrium is closest to:

  • 99.0%

  • 1.00%

  • 99.9 %

  • 0.100%

Solution

B.

1.00%

HA space leftwards harpoon over rightwards harpoon space straight H to the power of plus space plus space straight A to the power of minus straight K subscript straight a space space equals space fraction numerator left square bracket straight H to the power of plus right square bracket left square bracket straight A to the power of minus right square bracket over denominator left square bracket HA right square bracket end fraction space equals space fraction numerator left square bracket straight H to the power of plus right square bracket over denominator left square bracket HA right square bracket end fraction left square bracket straight H to the power of plus right square bracket space equals space square root of straight K subscript straight a left square bracket HA right square bracket end root space equals space square root of 1 space straight x space 10 to the power of negative 5 end exponent space 0.1 end root space equals square root of 1 space straight x 10 to the power of negative 6 end exponent end root space equals space 1 space straight x space 10 to the power of negative 3 end exponent straight alpha space equals space fraction numerator actual space ionization over denominator molar space concentration end fraction space equals space fraction numerator 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space 10 to the power of negative 2 end exponent percent sign space of space acid space dissociated space space equals space 10 to the power of negative 2 end exponent space straight x space 100 space equals space 1 percent sign

Question
CBSEENCH11008425
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Which of the following oxidation states are the most characteristics for lead and tin respectively? 

  • +4,+2

  • +2,+4

  • +4,+4

  • +2,+2

Solution

B.

+2,+4

The tendency to form +2 ionic state increase on moving down the group due to the inert pair effect.
Most characteristic oxidation state for lead and tin are +2,+4 respectively.

Question
CBSEENCH11008426
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The correct order of C-O bond length among CO, CO32-, CO2 is,

  • CO2<  CO32-<  CO

  • CO <  CO32-<  CO2

  • CO32- <CO2 <CO

  • CO < CO2 < CO32-

Solution

B.

CO <  CO32-<  CO2

The bond length is the average distance between the centres of nuclei of two bonded atoms. Centres of nuclei of two bonded atoms. A multiple bonds (double or triple bond) is always shorter than the corresponding single bond. 
The C- atom is CO32- is sp2 hybridised as shown

The C- atom is CO2 is sp hybridised with a bond distance of carbon -oxygen is 122 pm.
straight O equals straight C equals straight O space left right arrow space to the power of plus straight O space identical to space straight C minus straight O to the power of minus space left right arrow space straight O to the power of minus minus straight C identical to straight O to the power of plus
The C- atom in CO is sp hybridised with C-O bond distance is 110 pm:
So the correct order is 
CO <  CO32-<  CO2

Question
CBSEENCH11008427
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The following equilibrium constants are given:

N2 + 3H2  ⇌ 2NH3; K1

N2 +O2  ⇌ 2NO; K2

H2 + 1/2O2  ⇌ H2O' K3

The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:

  • K2K33 /K1

  • KK2K32 /K1

  • K22K3 /K1

  • K1K2 /K3

Solution

A.

K2K33 /K1

The required equation for the oxidation of NH3 by oxygen to give NO is:
4 NH subscript 3 space plus space 5 straight O subscript 2 space rightwards arrow from 800 to the power of straight o space straight C to Pt space left parenthesis gauze right parenthesis of space 4 space NO space plus space 6 straight H subscript 2 straight O For space this space space straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction For space the space equation space left parenthesis straight I right parenthesis space straight K subscript 1 space equals space fraction numerator left square bracket NH subscript 3 right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight H subscript 2 right square bracket cubed end fraction For space the space equation space left parenthesis II right parenthesis space straight K subscript 2 space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction For space the space equation space left parenthesis III right parenthesis space straight K subscript 3 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket over denominator left square bracket straight H subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction For space getting space the space straight K space we space must space do straight K subscript 1 squared space equals space fraction numerator left square bracket NH subscript 3 right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight H subscript 2 right square bracket to the power of 6 end fraction comma space straight K subscript 2 squared space equals space fraction numerator left square bracket NO right square bracket to the power of 4 over denominator left square bracket straight N subscript 2 right square bracket squared left square bracket straight O subscript 2 right square bracket squared end fraction straight K subscript 3 superscript 6 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket straight H subscript 2 right square bracket to the power of 6 left square bracket straight O subscript 2 right square bracket to the power of begin display style 6 over 2 end style equals 3 end exponent end fraction straight K space equals space fraction numerator straight K subscript 2 superscript 2 space straight x space straight K subscript 3 superscript 6 over denominator straight K subscript 1 superscript 2 end fraction space substituting space the space value space we space get comma straight K space equals space fraction numerator left square bracket NO right square bracket to the power of 4 left square bracket straight H subscript 2 straight O right square bracket to the power of 6 over denominator left square bracket NH subscript 3 right square bracket to the power of 4 left square bracket straight O subscript 2 right square bracket to the power of 5 end fraction space so space the space value space of space straight K space in space terms space of space straight K subscript 1 comma space straight K subscript 2 space and space straight K subscript 3 space is space space straight K thin space equals space fraction numerator straight K subscript 2 straight K subscript 3 superscript 3 over denominator straight K subscript 1 end fraction

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