Answer:
Question: The Haber-Bosch process involves the combination of nitrogen and hydrogen at high temperatures in the presence of a catalyst to produce ammonia. It is currently the main industrial source of ammonia.
Explanation:
Answer: The equilibrium concentration of CO at 1000 K is 0.016 M , the equilibrium concentration of at 1000 K is= 0.033 M and the equilibrium concentration of at 1000 K is 0.139 M
Explanation:
Initial concentration of
Initial concentration of
The given balanced equilibrium reaction is,
Initial conc. 0.1550 M 0.172 M 0 M
At eqm. conc. (0.1550-x) M (0.172-x) M (x) M
The expression for equilibrium constant for this reaction will be,
we are given :
Now put all the given values in this expression, we get :
Thus the equilibrium concentration of CO at 1000 K is= (0.1550-x) M =(0.1550-0.139) M = 0.016 M
Thus the equilibrium concentration of at 1000 K is= (0.172-x) M =(0.172-0.139) M = 0.033 M
Thus the equilibrium concentration of at 1000 K is= x M = 0.139 M
The H+ in a solution that has a Ph of 8.73 is calculate as follows
Ph is always = - log (H+)
H+ = 10^-Ph
H+ is therefore = 10 ^- 8.73
H+ = 1.86 x10^-9 M
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: