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 answer is volume because The base units of length and volume are linked in the metric system. By definition, a liter is equal to the volume of a cube exactly 10 cm tall, 10 cm long, and 10 cm wide. Because the volume of this cube is 1000 cubic centimeters and a liter contains 1000 milliliters, 1 milliliter is equivalent to 1 cubic centimeter.
Hope this helps :)
The angle- angle similarity postulate. It states that if a triangle has two equal corresponding angles the angles are similar.
Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, 
The conjugate base is
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.

Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
The liters in 3.25 g of ammonia 4.28 L
<u><em>calculation</em></u>
Step 1: find moles of ammonia
moles = mass÷ molar mass
From periodic table the molar mass of ammonia (NH₃) = 14 +(1×3 ) = 17 g/mol
3.25 g÷ 17 g/mol = 0.191 moles
Step 2: find the number of liters of ammonia
that is at STP 1 moles = 22.4 L
0.191 moles = ? L
<em>by cross multiplication</em>
={( 0.191 moles ×22.4 L) / 1 mole} = 4.28 L