The rate of particle movement decreases. A direct effect of this is the decrease in particle distance.
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:
pH= 9.2
Explanation:
Henderson hasselbach equation
pKa= log Ka= log (4.9 x 10^-10)=9.3
![pH=Pka+log \frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=pH%3DPka%2Blog%20%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
![pH=9.3+log \frac{[CN-]}{[HCN]}](https://tex.z-dn.net/?f=pH%3D9.3%2Blog%20%5Cfrac%7B%5BCN-%5D%7D%7B%5BHCN%5D%7D)
![pH=9.3+log \frac{[0.64 M]}{[0.83 M]}](https://tex.z-dn.net/?f=pH%3D9.3%2Blog%20%5Cfrac%7B%5B0.64%20M%5D%7D%7B%5B0.83%20M%5D%7D)
pH= 9.2
Answer:
K = 2.96x10⁻¹⁰
Explanation:
Based on the initial reaction:
N2O4 ⇄ 2NO2; K = 1.5x10³
Using Hess's law, we can multiply this reaction changing K:
3 times this reaction:
3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹
The inverse reaction has a K of:
6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;
<h3>K = 2.96x10⁻¹⁰</h3>
