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The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
Answer:

Explanation:
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In this case, since the applied current for the 50.0 mins provides the following charge to the system:

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

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Let the mass of CaO = x grams
So mass of BaO = 5.14 -x grams
moles of CaO = mass / molar mass = x / 56
Moles of BaO = mass / molar mass = 5.14-x / 153
initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06
final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018
So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles
x / 56 + (5.14-x) /153 = 0.042
on solving
x = 0.68
So mass of CaO = 0.68 g
So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %
Percentage of BaO = 86.6%