Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>
In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.
See figure 1
I hope it helps!
The organization is based off solitaire and many of the elements are the same
One mole is always the same number: 6.02 * 10^ 23.
So, one mole of cars = 6.02 * 10 ^23 cars; one mole of pencils = 6.02 * 10^23 pencils; one mole of atoms = 6.02 * 10^23 atoms; one mole of molecules = 6.02 * 10^23 molecules.
So, all the options are correct: one mole of calcium ions has 6.02 * 10^23 representative particles, such as one mole of calcium nuclei and one of calcium atoms.
It depends on what type of graph you have. The easiest would be using a H-T diagram. Enthalpy of vaporization is the physical change from liquid to vapor. It occurs at a constant pressure and a constant temperature. As shown in the picture, 1 point is drawn on the subcooled liquid, and another point of the saturated vapor isothermal line. Now, the difference between those two points is the value for the enthalpy of vaporization of water.
Answer:
D
Explanation:
To answer this question, we will need to write the dissociation equation of aluminum trichloride.
AlCl3 ——-> Al3+ + 3Cl-
It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion.
From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same
