An atomic nucleus having low n/p ratio tries to find stability by
the emission of an α-particle
the emission of a positron
capturing an orbital electron (K-electron capture)
emission of β -particle
B.
the emission of a positron
Emission of a positron results in the conversion of a proton into neutron, along with the release of neutrino. Thus, the number of protons decreases and of neutron increases, i.e., n/p ratio increases and the nuclei becomes stable.
Given that,
C + O2 → CO2 ; H = -x kJ
2CO + O2 → 2CO2 ;H = -y kJ
The heat of formation of carbon monoxide will be
y + 2x
2x - y
A.
Given,
C + O2 → CO2 ; = -x kJ ...(i)
2CO + O2 → 2CO2 ; = -y kJ ...(ii)
Required equation as per question is C + O2 → CO,
On reversing equation (ii), we get
2CO2 → 2CO + O2; = +y kJ ...(iii)
Dividing equation (iii) by 2 gives the following reaction
CO2 → CO + ; = + ...(iv)
On adding equation (i) and (iv), we get
C + → CO ;
In case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding molecular orbital resembles the character of A more than that of B. The statement
is false
is true
cannot be evaluated since data is not sufficient
is true only for certain systems
B.
is true
The given statement is true as electrons are shifted more towards the more electronegative atom.
(32Ge76, 34Se76) and (14Si30, 16S32) are examples of
isotopes and isobars
isobars and isotones
isotones and isotopes
isobars and isotopes
B.
isobars and isotones
32Ge76 and 34Se76 have the same mass number but different atomic number, therefore, they are isobars.
Number of neutrons, in
14Si30 = 30 - 14 = 16
16Si32 = 32 - 16 = 16
Because of the presence of same number pf neutrons, 14Si30 and 16S32, they are known as isotones.
The pH of 10-4M KOH solution will be
4
11
10.5
10
D.
10
10-4 M KOH = 10-4 M [OH-]
We know that,
[H+][OH-] = 1 × 10-14
[H+] × 10-4 = 1 × 10-14
[H+] = = 1 × 10-10 M
pH = - log [H+] = -log (1 × 10-10 M) = 10
The system that contains the maximum number of atoms is
4.25 g of NH3
8 g of O2
2 g of H2
4 g of He
C.
2 g of H2
Number of atoms = × NA × Number of atoms in 1 mole
Number of atoms in 8 gm
O2 =
Number of atoms in 2 gm
H2 =
Number of atoms in 4 gm
He =
Thus, 2 gm H2 contains the maximum number of atoms among the given.
The values of and of a certain reaction are -400 kJ mol-1 and -20 kJ mol-1K-1 respectively. The temperature below which the reaction is spontaneous, is
100 K
20°C
20 K
120°C
C.
20 K
Given,
H = -400 kJ mol-1 ; S = -20 kJ mol-1 K-1
Gibbs Helmholtez equation is
For a reaction to be spontaneous, G must be less than 0, i.e. it should be negative.
The value of H for cooling 2 mole of an ideal monoatomic gas from 225°C to 125°C at constant pressure will be [given Cp = ].
250 R
-500 R
500 R
-250 R
B.
-500 R
The value of H is
H = -nCp
= -2 × R ( 225 - 125)
= -5R (100)
= -500 R
Therefore, H of -500 R is required for cooling 2 mole of an ideal monoatomic gas from 225°C to 125°C at constant pressure.
98Cf246 was formed along with a neutron when an unknown radioactive substance was bomabrded with 6C12. The unknown substance is
91Pa234
90Th234
92U235
92U238
C.
92U235
Supoose, the unknown radioactive substance is xAy.
xAy + 6C12 → 98Cf246 + 0n1
On comparing atomic numbers, we get
x + 6 = 98 + 0
x = 98 - 6 = 92
On comparing mass numbers, we get
y + 12 = 246 + 1
y = 247 - 12 = 235
Since, uranium has the atomic number 92, thus, the given unknown substance was 92U235.
The enthalpy of vaporisation of a certain liquid at its boiling point of 35°C is 24.64 kJ mol-1. The value of change in entropy for the process is
704 JK-1mol-1
80 JK-1mol-1
24.64 JK-1mol-1
7.04 JK-1mol-1
B.
80 JK-1mol-1
Entropy of vapourisation (S) =
Given, H = 24.64 kJ mol-1 and boiling point = 35 + 273 = 308 K