Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
Offspring are more rapidly reproduced
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the<em> PV=nRT formula</em>, where:
- R = 8314.46 Pa·L·mol⁻¹·K⁻¹
We<u> input the data given by the problem</u>:
- 205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 K
And <u>solve for n</u>:
'Acceleration' means any change in speed or direction
of motion.
so
C). No acceleration. Straight, at constant speed.
No change of speed or direction.
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
