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10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

P1=1 atm

          P2= 1/2=0.5 atm

          T1=273.15 K

          V2=?

          V1=?

32 g dioxygen occupies = 22.4 L volume at STP

∴ 1.6 g dioxygen will occupy = 22.4L x 1.6g / 32g = 1.12 L

     V1=1.12 L

From Boyle's law (as temperature is constant)

p1V1=p2V2

V2=p1V1p2

    = 1 atm x 1.12 l/0.6 atm g = 2.24 L

 

(ii) Number of molecules of dioxygen.

  

...more

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The line spectrum associated with any element possesses lines corresponding to specific wavelengths and these lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy i.e., quantized values.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: As per de Broglie, every object in motion has a wave character i.e. every object/matter have dual nature both particle and wave nature. But the wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.

Given, m (mass) = 100g, v=100km/hr

 We have

λ = h/mv= 238.5 * 10-34 m

As the wavelength associated with the cricket ball is so small that it can't be detected.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: We have V = 109677 [1/ni 2 - 1/nf 2 ] cm-1

 Given, ni = 2, nf=4

V =109677 [1/4 - 1/16] = 20564.44 cm-1

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: As per the Hund's rule the half filled and fully filled orbital leads to the extra stability due to the symmetry thus fully filled 3d and half filled 4s is preferred.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans:  

We know that energy is inversely proportional to the wavelength

Thus, the increasing order of energy is B

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The mass number (A) is defined as the sum of the number of protons and neutrons present in the nucleus and the atomic number is defined as the number of protons or electrons present in an atom. Thus A=13, the number of neutrons is 7 so the number of protons is 6. Thus, the atomic number is 6.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The charged particles get deflected by the electric field. As among the given option only neutron is the particle that is neutral thus it does not get deflected by the electric field.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: Based upon the above information, arrange the following orbitals in the increasing order of energy.

(a) 1s, 2s, 3s, 2p

(b) 4s, 3s, 3p, 4d

(c) 5p, 4d, 5d, 4f, 6s

(d) 5f, 6d, 7s, 7p

     Thus, the increasing order of energy is 3s<3p<4s<4d

Thus, the increasing order of energy is 4d<5p<6s<4f<5d

Thus, the increasing order of energy is 7s<5f<6d<7p

(II) Based upon the above information, solve the questions given below :

(a) Which of the following orbitals has the lowest energy?

4d, 4f, 5s, 5p

Ans -

           The lowest

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New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The number of radial nodes is given by n-l-1, where n is principal quantum number, l is azimuthal quantum number. The number of angular nodes is given by n-l, where n is principal quantum number, l is azimuthal quantum number. Here n =3 and l =1 Thus, angular nodes = 3-1 = 2 and radial node = 3-1-1 = 1.

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