A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl d
1 answer:
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Answer:
Δ
Δ
Explanation:
Hello,
At the first state, the molar volume is:
The volume in both the second and third state:
Now, as it is about an adiabatic process, one remembers the following relationships:
- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:
- And the temperature:
- Q:
It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:
Then the total heat:
- The work for the first process is:
It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:
- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:
- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:
Best regards.
Explanation:
The given data is as follows.
= 253.4 nm =
(as 1 nm =
)
= 5,
= ?
Relation between energy and wavelength is as follows.
E =
=
= J
=
Hence, energy released is .
Also, we known that change in energy will be as follows.
where, Z = atomic number of the given element
0.02 + 0.04 =
=
= 4
Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.