There are several ways to visually represent compounds. For this particular organic compound, we can use the skeletal formula and the expanded formula. The skeletal makes use of lines to show which atoms are bonded to each other. The expanded formula shows the species of the atoms and their bonding with other atoms. I have attached the two representations.
Answer:
C. 2.000 M C6H12O6
Explanation:
Let us obtain the molarity of the solution.
Molar Mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol
Mass of C6H12O6 = 180g
Number of mole = Mass /Molar Mass
Number of mole of C6H12O6 = 180/180 = 1mole
Volume = 500mL = 500/1000 = 0.5L
Molarity = mole /Volume
Molarity = 1/0.5
Molarity = 2M
So the solution will be best labelled as 2M C6H12O6
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
