Class 12th
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New answer posted
a year ago
Contributor-Level 10
gives iodoform test and slow Lucas test, so it's a methyl secondary alcohol. D gives fast Lucas test, so it's a tertiary alcohol. The Grignard products must be tertiary and secondary alcohols. So A and B must be an aldehyde and a ketone.
New answer posted
a year agoContributor-Level 10
. [XeF? ]? has 7 electron pairs (5 bonding, 2 lone), pentagonal planar. XeO? F? has 5 electron pairs (trigonal bipyramidal).
New answer posted
a year agoContributor-Level 9
Sol. (x/m) = k (P)¹/?
log (x/m) = logk + 1/n logP
Slope = 1/n = 2 So n = 1/2
Intercept ⇒ logk = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3² = 48
New answer posted
a year agoContributor-Level 9
The oxidation states of iron in these compounds will be
A = +2
B = +4
C = 0
The sum of oxidation states will be = 6.
New answer posted
a year agoContributor-Level 9
Sol. E? cell = E? (Sn²? |Sn) - E? (Cu²? |Cu)
= -0.16 - 0.34 = -0.50V
ΔG? = -nFE? cell
= -2 * 96500 * (-0.5) = 96500 J
= 96.5 kJ = 96500 J
New answer posted
a year ago
Contributor-Level 10
Reaction (1) is SN1 (rate independent of [OH? ]). Reaction (2) is E2 (rate depends on [OH? ]).
Statement (B) is correct. Changing concentration of base will have no effect on reaction (1).
New answer posted
a year agoContributor-Level 9
Cesium has lowest ionisation enthalpy and hence it can show photoelectric effect to the maximum extent hence it is used in photo electric cell.
New question posted
a year ago
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