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NEET chemistry

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Question
CBSEENCH12011009
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What is the maximum number of orbitals that can be identified with the following quantum numbers?

n=3, l =1, m1 = 0

  • 1

  • 2

  • 3

  • 4

Solution

A.

1

The value of n=3 and l =1 suggest that it is a 3p orbital while the value of m1 = 0 [magnetic quantum number] shows that the given 3p orbital is 3pz in nature.
Hence, the maximum number of orbitals identified by the given quantum number is only 1, i.e. 3pz.

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Question
CBSEENCH12011010
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Calculate the energy in joule corresponding to light of wavelength 45mm: (Planck's constant h=6.63 x10-34)Js; speed of light c= 3 x 108 ms-1)

  • 6.67 x 1015

  • 6.67 x 1011

  • 4.42 x 10-15

  • 4.42 x 10-18

Solution

D.

4.42 x 10-18

The wavelength of light is related to its energy by the equation,
straight E space equals hc over straight lambda Given comma straight lambda space equals space 45 space mm space equals space 45 space straight x space 10 to the power of negative 9 end exponent space straight m space left square bracket therefore space 1 space nm space equals space 10 to the power of negative 9 end exponent space straight m space right square bracket Hence comma space straight E equals fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent space js space straight x space 3 space straight x space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 45 space straight x space 10 to the power of negative 9 end exponent space straight m end fraction space equals space 4.42 space straight x space 10 to the power of negative 18 end exponent space straight j Hence comma space the space energy space corresponds space to space light space of space wavelenght space 45 space nm space is space 4.42 space straight x space 10 to the power of negative 18 space end exponent space straight j

Question
CBSEENCH12011011
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Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 270 C in identical conditions. The ratio of the volumes of gases H2 : O2: CH4 would be 

  • 8:16:1

  • 16:8:1

  • 16:1:2

  • 8:1:2

Solution

C.

16:1:2

According to Avogadro's hypothesis,
Volume of a gas (V)  straight alpha number of moles (n)
Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H2:O2: methane (CH4) is given by
straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight n subscript straight H subscript 12 end subscript space colon straight n subscript straight O subscript 2 end subscript space colon space straight n subscript CH subscript 4 end subscript rightwards double arrow space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight m subscript straight H subscript 2 end subscript over straight M subscript straight H subscript 2 end subscript colon space straight m subscript straight O subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript space colon straight m subscript CH subscript 4 end subscript over straight M subscript CH subscript 4 end subscript But space straight m subscript straight H subscript 2 end subscript space equals space straight m subscript straight O subscript 2 end subscript space equals space straight m subscript CH subscript 4 end subscript space equals space straight m space open square brackets therefore space straight n space equals space fraction numerator mass over denominator molar space massd end fraction close square brackets Thus comma space space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space space equals space straight m over 2 space equals space straight m over 1 space equals straight m over 16 space equals space 16 colon space 1 colon 2

Question
CBSEENCH12011012
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Which of the following salts will give highest pH in water?

  • KCl

  • NaCl

  • Na2CO3

  • CuSO4

Solution

C.

Na2CO3

The highest pH refers to the basic solution containing OH- ions. Therefore, the basic salt releasing OH- ions on hydrolysis will give highest pH in water.
Only the salt of a strong base and weak acid would release OH-ion on hydrolysis. Among the given salts, Na2CO3 corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H2CO3 [weak acid].
CO subscript 3 superscript 2 minus end superscript space plus straight H subscript 2 straight O space rightwards harpoon over leftwards harpoon HCO subscript 3 superscript minus space plus OH to the power of minus

Question
CBSEENCH12011013
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When 22.4 L of H2 (g) is mixed with 11.2 L of Cl2 (g), each of at STP, the moles of HCl (g) formed is equal to

  • 1 mole of HCl (g)

  • 2 moles of HCl (g)

  • 0.5 mole of (g) 

  • 1.5 mole of HCl(g)

Solution

A.

1 mole of HCl (g)

The given problem is related to the concept of the stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then identify the limiting reagent [possessing minimum number of moles and completely used up in the reaction]
The limiting reagent gives the moles of product formed in the reaction.
space space space space space space space space space space space space space space space space space space space space straight H subscript 2 space left parenthesis straight g right parenthesis space plus space space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis initial space vol. space space 22.4 space straight L space space space space space space space space 11.2 space straight L space space space space space space 2 space mol therefore space 22.4 space straight L space volume space at space STP space is space occupied space by Cl subscript 2 space equals space 1 space mole therefore space 11.2 space straight L space volume space will space be space occupied space by comma Cl subscript 2 space equals space fraction numerator 1 space straight x space 11.2 over denominator 22.4 end fraction space mole space equals space 0.5 space mol Thus comma stack space straight H subscript 2 space with 1 mole below left parenthesis straight g right parenthesis space plus stack Cl subscript 2 with 0.5 below space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis Since comma Cl subscript 2 space possesses space minimum space number space of space moles comma thus space it space is space the space limiting space reagent. As space per space equation comma 1 space mol space Cl subscript 2 space identical to space 2 space mol space HCl therefore space 0.5 space mol space Cl subscript 2 space equals space 2 space straight x 0.5 space mol space HCl Hence comma space 1.0 space mole space of space HCl space left parenthesis straight g right parenthesis space is space priduced space by space 0.5 space mole space of space Cl subscript 2.

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