Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
Answer: the reaction rate of the forward reaction increases (although it is not among the options shown).
Explanation:
1) The equation given is:
CO (g) + Cl₂ (g) ⇄ COCl₂ (g)
2) The forward reaction is:
CO (g) + Cl₂ (g) → COCl₂ (g)
3) The stoichiometry is 1 mole + 1 mole → 1 mole
4) Analysis and conclusion:
The effect of change in the conditions of an equilibrium is foreseen using Le Chatelier's principle.
Le Chatelier's principle states that a system in equilibrium will act trying to compensate the distress applied to the system.
In this case, the distress is increase in pressure.
To compensate the increase in pressure the system must act by reducing the number of moles in the system. Since the forward reaction implies the conversion of two moles of reagents into one mole of product, the more the forward reaction progress the less the number of moles will be present in the system and so the more the compensation of the increase in pressure. So the forward raction is favored by the increase in pressure.
Favoring the forwar reaction means that its rate will incrase,
Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2
According to Nernst equation for electrochemical cell,
Ecell = -2.303
![\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}](https://tex.z-dn.net/?f=%20%5Cfrac%7BRT%7D%7BnT%7D%20log%20%20%5Cfrac%7B%5BZn%5E2%2B%5DR%7D%7B%5BZn%5E2%2B%5DL%7D%20%20)
= 0.014
Given: T =
, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v
∴ 0.014 = - 2.303
∴ log
=
= -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m
Answer:
Too high a value
Explanation:
HA + NaOH ⟶ NaA +H₂O
If the student has gone slightly past the equivalence point, they have added too much base.
The moles of HA are directly proportional to the moles of NaOH, so the moles of acid that the student calculates will be too high.
The calculated concentration of acid will also be too high.
