Answer:
V₂ = 12.43 L
Explanation:
Given data:
Initial pressure = 650 KPa
Initial volume = 2.2 L
Final pressure = 115 KPa
Final volume = ?
Solution:
The given problem will be solved through the Boyles law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
650 KPa ×2.2 L = 115 KPa × V₂
V₂ = 1430 KPa. L/ 115 KPa
V₂ = 12.43 L
Answer:
Approximately 56.8 liters.
Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.
Explanation:
Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words
,
where
is the final volume,
is the initial volume,
is the final temperature in degrees Kelvins.
is the initial temperature in degrees Kelvins.
Convert the temperatures to degrees Kelvins:
.
.
Apply Charles's Law to find the new volume of this gas:
.
Answer:
SN2
Explanation:
The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.
The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.
Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.
Answer:False
Explanation:
slow down/ meet and obstacle.
Answer:
Option 5.
Equal amounts of hydroxide and hydronium ions
Explanation:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw: 1×10⁻¹⁴
Neutral solution has pH = 7
So, [H₃O⁺] = [OH⁻]
In solutions of only pure water:
[H₃O⁺] = 1×10⁻⁷ M
[OH⁻] = 1×10⁻⁷ M
Kw = [H₃O⁺] . [OH⁻] → 1×10⁻⁷ . 1×10⁻⁷ = 1×10⁻¹⁴