The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the following Charle's law equation:
V1/T1 = V2/T2
Where;
- T1 = initial temperature
- T2 = final temperature
- V1 = initial volume
- V2 = final volume
- V1 = 9L
- V2 = ?
- T1 = 325K
- T2 = 450K
9/325 = V2/450
325V2 = 4050
V2 = 4050/325
V2 = 12.46L
Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
Learn more about volume at: brainly.com/question/2817451
Explanation:
It is given that r = 0.283 nm. As 1 nm = 
.
Hence, 0.283 nm = 
- Formula for coulombic energy is as follows.
where, e = 
C
= 
= 
- As 1 eV =
So, 1 J = 
Hence, U = 
= 8.9 eV
- Also, 1 J =
= 
kJ/mol
Therefore, U = 
kJ/mol
= 
Answer:
The metric system goes by powers of ten, so it's very easy to measure. That would be the main advantage, measurements of ten. We can also say it's the most used measurement around the world, so all scientists have little to no conversion, but the main answer is probably the first one :)
Answer:
Explanation:
Discussion
When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?
N2(g) + 3H2(g) ⇌ 2NH3(g)
The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols
on the left side.
The right side has 2 mols of NH3 = 2 mols on the right.
Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.
Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.
2NH3 is only two mols.
The equilibrium shifts Right
Answer
D
N2H4
H3CO2H
C2H2O4
This should be the right answer, let me know if its not so i can know for future.
