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!
Answer:
30.8 g of water are produced
Explanation:
First of all we need the equation for the production of water:
2H₂ + O₂ → 2H₂O
2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.
As we assume, the oxygen in excess, we determine the moles of H₂.
1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles
Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water
Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
We know that,
The relation between the 
for an ideal gas are :
As we are given :
Now we have to calculate the entropy change of the gas.
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
D = m / V
13.6 = 8.3 / V
V = 8.3 / 13.6
V = 0.610 mL
hope this helps!
