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The value of Planck's constant is 6.63 x 10-34 Js. The speed of light is 3 x1017 nms-1 . Which value is closet to the wavelength in nanometer of a quantum of light with frequency of 6 x 1015 s-1 ?
10
25
50
75
C.
50
Given, Planck's constant,
h= 6.63 x10-34
speed of light, c= 3 x1017 nms-1
Frequency of quanta
v=6 x1015 s-1
Wavelength, λ =?
We know that,
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What is the maximum number of electrons that can be associated with the following set of quantum number?
n=3, l =1 and m=-1.
10
6
4
2
D.
2
The orbital of the electron having =3, l =1 and m= -1 is 3pz (as nlm) and an orbital can have a maximum of two electrons with opposite spins.
therefore, 3pz orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.
Based on equation
E=-2.178 x 10-18 J 
certain conclusions are written. Which of them is not correct?
The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than if would be if the electrons were at the infinite distance from the nucleus.
Larger the value of n, the larger is the orbit radius
Equation can be used to calculate the change in energy when the electron changes orbit
For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
D.
For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
If n=1,
E1 = - 2.178 x 10-18 Z2 J
If n=6
From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. it means that electron is more strongly bound in the smallest allowed orbit.
How many grams of the concentrated nitric solution should be used to prepare 250 mL o 2.0M HNO3 ? The concentrated acid is 70% HNO3.
45.0 g conc. HNO3
90.0 g conc. HNO3
70.0 g conc. HNO3
54.0 g conc. HNO3
A.
45.0 g conc. HNO3
Given, molarity of solution = 2
Volume of solution = 250 mL = 250/1000 = 1/4 L
Molar mass of
HNO3 = 1+14+3 x 16 = 63 g mol-1
therefore, Molarity
= 
Dipole-induced dipole interactions are present in which of the following pair?
H2O and alcohol
Cl2 and CCl4
HCl and He atoms
SiF4 and He atoms
C.
HCl and He atoms
Dipole-induced dipole interaction is present in the pair in which the first species is polar and the other is non-polar.
H2O and alcohol both are non-polar so there exist dipole-dipole interactions in between them.
Cl2 and CCl4 both are non-polar so there exists induced dipole -induced dipole interactions in between them. Similarly is true for SiCl4 and He atoms pair.
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