Answer:
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Explanation:
Here, 
H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation
H(diene) depends on stability of diene.
More stable a diene, lesser will be it's H(diene) value (more neagtive).
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Hence, 
is higher (less negative) for trans-1,3-pentadiene
Explanation:
A. Hydrogen bonding is present in CS2 but not in CO2.
B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.
C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.
D. The dispersion forces are greater in CS2 than in CO2.
<u><em>PLS MARK BRAINLIEST :D</em></u>
Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
Answer:
A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.
Explanation:
We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.
In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.
Answer:
Explanation:
Using freezing point depression formula,
ΔTemp.f = Kf * b * i
Where,
ΔTemp.f = temp.f(pure solvent) - temp.f(solution)
b = molality
i = van't Hoff factor
Kf = cryoscopic constant
= 1.86°C/m for water
= (0 - (-5.58))/1.86
= 3.00 mol/kg
Assume 1 kg of water(solvent)
= (3.00 x 1)
= 3.00 mol.
