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
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
Basis: 100 mL solution
From the given density, we calculate for the mass of the solution.
density = mass / volume
mass = density x volume
mass = (1.83 g/mL) x (100 mL) = 183 grams
Then, we calculate for the mass H2SO4 given the percentage.
mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams
Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles
Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M
Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
Answer:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
Explanation:
Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.
Our compound is the RbOH.
When it is put in water, i can dissociate like this:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
Answer:
Her oxygen would be traveling throughout her body and carbon dioxide will be released every time she breathes out. The oxygen will come in through the respiratory system which is a body system that helps you breathe with lungs, which then delivers the oxygen molecules to other organs and body systems. Oxygen will go through the circulatory system and circle around until the some oxygen goes in the cells or the stored glucose. Then some Oxygen molecules go into the cells and make energy with Glucose. She would have higher glucose/ healthy sugar to be considered healthy. But if she was healthy she would have glucose in her digestive system( where the food goes when you swallow it some of the parts are the stomach and intestines both small and large) which will be transferred to the circulatory system then it will circle around the system
Explanation: