Answer:
A
Explanation:
because your body needs to have an intake in the volume of liquids, and when you sweat it secretes the water from your body
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
Learn more: brainly.com/question/2510654
Answer:
We will derive the combined gas equation from the law of Gases
Boyle's law
P∝1/V
PV=Constant
Charles law
V∝T
Avogadro's Law
V∝n
By combining these three equations we get the combined gas equation
PV=nRT
V=nRT/P
If n=1 mole
V=RT/P
By putting the value of R, T, and P in the above equation we can calculate the volume of the gas at STP.
Answer:
6.5x10⁻³M = [OH⁻]
Explanation:
The Kb of a Weak base as ethylamine is expressed as follows:
Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]
As the equilibrium of ethylenamine is:
C₂H₅NH₂(aq) + H₂O(l) ⇄ C₂H₅NH₃⁺(aq) + OH(aq)
The concentration of C₂H₅NH₃⁺(aq) + OH(aq) is the same because both ions comes from the same equilibrium. Thus, we can write:
Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]
6.4x10⁻⁴ = [X] [X] / [C₂H₅NH₂]
Also, we can assume the concentration of ethylamine doesn't decrease. Replacing:
6.4x10⁻⁴ = [X] [X] / [0.066M]
4.224x10⁻⁵ = X²
6.5x10⁻³M = X
<h3>6.5x10⁻³M = [OH⁻]</h3>
