C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
Answer:
Time take to deposit Ni is 259.02 sec.
Explanation:
Given:
Current 
A
Faraday constant 
Molar mass of Ni 
Mass of Ni 
g
First find the no. moles in Ni solution,
Moles of Ni 
mol
From the below reaction,
⇆ 
Above reaction shows "1 mol of 
requires 2 mol of electron to form 1 mol of "
So for finding charge flow in this reaction we write,
Charge flow 
C
For finding time of reaction,
Where 
charge flow
sec
Therefore, time take to deposit Ni is 259.02 sec.
Answer:
(3S)-2-chloro-2,3-dimethylpentane is produced exclusively.
Explanation:
Electrophilic addition to (3S)-2,3-dimethylpent-1-ene proceeds through a carbocationic intermediate.
In the first step, 
adds onto double bond to produce more stable tertiary carbocation. (protonation)
In the second step, 
adds onto carbocation to produce (3S)-2-chloro-2,3-dimethylpentane exclusively.(nucleophilic addition)
So, option (d) is correct.
Answer:
24.24 L
Explanation:
Boyle’s law, also called Mariotte’s law, a relation concerning the compression and expansion of a gas at constant temperature.
This empirical relation, formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant.
Real gases obey Boyle’s law at sufficiently low pressures, although the product pv generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.
As, PV = k
P₁ V₁ = P₂ V₂
Given P₁ = 101 KPa
V₁ = 6 L
P₂ = 25 kPa
So, V₂ = P₁ V₁ /P₂ = 101 *6/25 = 24.24 L
Answer: The transition elements are in the d-block, and in the d-orbital have valence electrons. They can form several states of oxidation and contain different ions. Inner transition elements are in the f-block, and in the f-orbital have valence electrons.
Explanation:
