At what temperature would a gas have a volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a pressure of 0.612 atm and a temperature of 28 C
Answers
Answer : The initial temperature of gas would be, 269.7 K
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 0.723 atm
= final pressure of gas = 0.612 atm
= initial volume of gas = 13.5 L
= final volume of gas = 17.8 L
= initial temperature of gas = ?
= final temperature of gas =
Now put all the given values in the above equation, we get:
Thus, the initial temperature of gas would be, 269.7 K
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Answers
Answers
Step 1:
Epoxidation of Alkene:
In first step Cyclohexene is treated with peroxybenzoic acid in the presence of Dichloromethane which results in the formation of epoxide moiety.
Step 2:
Nucleophillic Substitution Reaction:
In second step the epoxide formed is treated with Sodium Methoxide which acts as a nucleophile and attacks the less hindered carbon of epoxide resulting in the formation of final product as shown below.
Final answer:
The best reagent and condition for a chemical reaction is dictated by the reactants at hand and the substance that you're trying to produce. For instance, to convert an alkene into an alcohol, Osmium Tetroxide (OsO4) at room temperature would be an example of an ideal reagent and condition.
Explanation:
In chemistry, when you are asked to place the best reagent and conditions in a reaction box, you are trying to predict the proper chemical, heat, or pressure conditions that will foster a certain chemical reaction. This requires understanding of substances' chemical properties, behavior under different conditions, and reaction mechanisms. For instance, if we want to oxidize an alkene into an alcohol, we would choose a reagent like Osmium Tetroxide (OsO4). In this case, OsO4 would be our 'best reagent', and room temperature may serve as the ideal condition as it generally facilitates this process. Each reagent and condition depend on the reactants you start with and the product you want at the end.
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Answers
Answer:
-1215.9J is the work done
Explanation:
It is possible to find work done in the change of volume of a gas at constant pressure using:
W = -P×ΔV
Where W is work, P is pressure and V is change in volume}
Replacing:
W = -6atm×(5L-3L)
W = -12atmL
As 1atmL = 101,325J, work done in joules is:
-12atmL ×(101.325J / atmL) = -1215.9J is the work done
The value of work done when a volume increases from 3 liters to 5 liters at 6 atm of pressure is -1215.9 Joules.
How do we calculate workdone?
Workdone on any boby or by the body will be calculated as:
W = -P×ΔV, where
W = workdone
P = applied or exerted pressure = 6 atm
ΔV = change in volume due to workdone = (5-3) L
Negative sign in the formula shows that work is done on the opposite side of the pressure or volume.
On putting all these values on the above equation, we get
W = -6atm × (5L-3L)
W = -12 atmL
We know that, 1 atmL = 101,325 J,
So, workdone in joules will be written as:
-12 atmL × (101.325J / atmL) = -1215.9 J
Hence, -1215.9 J is the workdone.
To know more about workdone, visit the below link:
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Answers
Answers
Final answer:
Among the given figures, 2.5 x 10-2 L (or 25 mL when converted to milliliters) is the smallest volume.
Explanation:
The student is asked to compare and determine the smallest volume between 2500 mL, 250 cm3, 2.5 x 10-2 L, and 25 m3. First, let's convert all volumes into a common unit, the milliliter (mL).
- 2500 mL = 2500 mL (already in mL),
- 250 cm3 = 250 mL (since 1 cm3 equals 1 mL),
- 2.5 x 10-2 L = 0.025 L = 25 mL (since 1 L equals 1000 mL),
- 25 m3 = 25 x 1,000,000 L = 25 x 1,000 x 1,000,000 mL = 25,000,000,000 mL.
From this comparison, we can see that 2.5 x 10-2 L or 25 mL is the smallest volume among the given figures.
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