If a gas expands at constant temperature indicates that
kinetic energy of molecules decreases
pressure of the gas increases
kinetic energy of molecules remains as same
number of the molecules of gas increase
C.
kinetic energy of molecules remains as same
Here, the temperature of gas remains constant hence kinetic energy of molecules remains the same.
If n = 6, the correct sequence for filling of electrons will be
ns (n - 1)d (n - 2)f np
ns (n - 2)f np (n - 1)d
ns np (n - 1)d (n - 2)f
ns (n - 2)f (n - 1)d np
D.
ns (n - 2)f (n - 1)d np
For n = 6 electron are filled as
6s 4f 5d 6p
This is because electron first enters in that orbitals for which (n + 1) is lower. In case, if (n + 1) is same, it goes first in that orbital for which n is lower.
An electron is moving in Bohr's fourth orbit. Its de-Broglie wave length is . What is the circumference of the fourth orbit?
2
4
C.
4
According to Bohr's concept, an electron always move in the orbit with angular momentum (mvr) equal to
For the reaction, N2(g) + O2(g) 2NO (g), the equilibrium constant is K1. The equilibrium constant is K2 for the reaction, 2NO(g) + O2)g) 2NO2(g). What is K for the reaction, NO2(g) ?
B.
N2(g) + O2(g) 2NO(g); K1 ...(i)
2NO(g) + O2(g) 2NO2(g); K2 ...(ii)
N2(g) + 2O2(g) 2NO2(g); K = K1 K2
For NO2(g) N2(g) + O2(g); K =
Among the following the molecule possessing highest dipole moment is
CO2
BF3
SO2
Trans-2-butene
C.
SO2
Among the given compounds, SO2 has unsymmetrical or angular structure. Thus, it has the highest dipole moment.
The angular shape of ozone molecule (O3) consists of
1 sigma and 2 pi bonds
2 sigma and 2 pi bonds
1 sigma and 1 pi bonds
2 sigma and 1 pi bonds
D.
2 sigma and 1 pi bonds
Single bond contains= 1 bond
Double bond contains = 1 bond + 1 bonds
Triple bond contains= 1 bond + 2 bonds
The molecule of O3 is bent with an angle 116.8° and equal O - O distance of 128 pm. The ozone molecule consists of 2 sigma and 1 pi bonds. The angular shape of molecule is represented as
The correct order of magnitude of bond angles among the compounds CH4, NH3 and H2O is
CH4 > H2O < NH3
H2O < NH3 < CH4
NH3 < CH4 < H2O
NH3 < H2O < CH4
B.
H2O < NH3 < CH4
Bond angles depend upon number of lone pair (s) of electrons. Higher the number of lone pairs of electrons, lesser the bond angle.
The dissociation equilibrium of a gas AB2 can be represented as
2AB2 (g) 2AB (g) + B2(g)
The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant KP and total pressure p is
B.
Initial moles 1 0 0
At equillibrium 2(1-x) 2x x
where, x = degree of dissociation
Total moles at equilibrium = 2 - 2x + 2x + x
= (2 + x)
So,
A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of N is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1. 8 x 10-5, what is the pH of this solution? (log 2. 7 =0. 43).
9.43
11.72
8.73
9.08
A.
9.43
pOH = pKb +
=
=
= 5 - 0.25 + (-0.176)
= 4.75 - 0.176 = 4.57
pH = 14 - 4.57 = 9.43
Which of the following is least likely to behave as Lewis base?
NH3
BF3
OH-
H2O
B.
BF3
Boron in BF3 is electron poor or electron deficient species. It has an empty orbital, so it can accept a pair of electrons, therefore, making it a lewis acid.
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