The answer for the following question is explained.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1).</em></u>
Explanation:
Here;
n represents the principal quantum number
l represents the Azimuthal quantum number
m represents magnetic quantum number
s represents spin quantum number
n = 5,
l = 2,
m = -2,
s = +1/2
Here, it implies 5d orbital.
In the 5d orbital, 10 electrons.
As the magnetic quantum number is -2, and so it can have 1 electron.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1)</em></u>
IT forms because they are highly reactive elements.
Answer:
4 lead = Pb
2 nitrogen = N
6 oxygen = O
Explanation:
Know the rules of multiplying wth perentheses.
- Hope that helped! Please let me know if you need further explanation.
Answer:
The standard cell potential is 2.00 V
Explanation:
<u>Step 1:</u> Data given
Cu is cathode because of
higher EP
Al3++3e−→Al E∘=−1.66 V anode
Cu2++2e−→Cu E∘=0.340 V cathode
<u>Step 2:</u> Balance both equations
2*(Al → Al3+-3e−) E∘=1.66 V
3*(Cu2++2e−→Cu) E∘=0.340 V
<u>Step 3:</u> The netto equation
2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-
2 Al + 3Cu2+ → 2Al3+ + 3Cu
<u>Step 4:</u> Calculate the standard cell potential
E∘cell = E∘cathode - E∘anode
E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al
E∘cell =0.340 V - (-1.66) = 2.00 V
The standard cell potential is 2.00 V
