In diborane, the two H - B - H angles are nearly
60°, 120°
95°, 120°
95°, 150°
120°, 180°
B.
95°, 120°
In diborane, the two H - B - H angles are nearly 95°, 120°. The structure can be represented as -
When 10 mL of 0.1 M acetic acid (pKa = 5.0) is titrated against 10 ml of 0.1 M ammonia solution (pKb = 5.0), the equivalence point occurs at pH
5.0
6.0
7.0
9.0
C.
7.0
pKa = -logKa, pKb = -logKb
F2 formed by reacting K2MnF6 with
SbF5
MnF3
KSbF6
MnF4
A.
SbF5
F2 is formed by reacting K2MnF6 with SbF5.
K2MnF6 + 2SbF5 → 2KSbF6 + MnF3 + F2
Among the following molecules
(i) eO3 (ii) eOF4 (iii) eF6
those having same number of lone pairs on Xe are
(i) and (ii) only
(i) and (iii) only
(ii) and (iii) only
(i), (ii) and (iii)
D.
(i), (ii) and (iii)
Among all the given options, option d is correct. All the 3 images have same number of lone pairs on Xe. Their structure are as follows-
XeO3 -
XeOF4 -
XeF6 -
The diamagnetic species is
[Ni(CN)4]2-
[NiCl4]2-
[CoCl4]2-
[CoF6]2-
A.
[Ni(CN)4]2-
Among all the given options, option a is correct. [Ni(CN)4]2- is a dimagnetic species. [NiCl4]2-, [CoCl4]2-, [CoF6]2 are paramagnetic species.
An endothermic reaction with high activation energy for the forward reaction is given by the diagram
C.
Endothermic reactions are those which involve absorption of heat. High activation energy means potential energy of product must be much greater than reactants.
The isoelectronic pair is
Cl2O, ICl2-
ICl2-, ClO2
IF2+, I3-
ClO2-, ClF2+
D.
ClO2-, ClF2+
Among all the given options, isoelectronic pair is ClO2-, ClF2+.
Number of electrons in ClO2- = 7 + 6 + 6 + 1 = 20
Number of electrons in ClF2+ = 7 + 7 + 7 - 1 = 20
(298 K) of methanol is given by the chemical equation
CH4 (g) + O2 (g) →CH3OH(g)
C (graphite) + O2 (g) + 2H2 → CH3OH(l)
C (diamond) + O2(g) + 2H2(g) → CH3OH(l)
CO(g) + 2H2 → CH3OH(l)
B.
C (graphite) + O2 (g) + 2H2 → CH3OH(l)
Methanol can be prepared synthetically by heating carbon monoide and hydrogen gases under pressure in the presence of a catalyst.
C (graphite) + O2(g) → CO(g) ...(i)
CO(g) + 2H2(g) → CH3OH(l) ...(ii)
From (i) and (ii)
C (graphite) + O2(g) + 2H2(g) → CH3OH(l)
emits 8 α-particles and 6 β-particles. The neutron/proton ratio in the product nucleus is
60/41
61/40
62/41
61/42
C.
62/41
The neutron/ proton ratio in the product nucleus is 62/41. 92U238 82X206
Number of protons = 82
Number of neutrons = 124
Neutron/proton ratio in the product nucelus =
α-Particles can be detected using
thin aluminium sheet
barium sulphate
zinc sulphide screen
gold foil
C.
zinc sulphide screen
α-particles can be detected by using zinc sulphide screen. It is used as phosphor in detection.
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