This
electronic transition would result in the emission of a photon with the highest
energy:
4p
– 2s
<span>This
can be the same with the emission of 4f to 2s which would emit energy in the
visible region. The energy in the visible region would emit more energy than in
the infrared region which makes this emission to have the highest energy.</span>
Orbital diagram:
<h3>Explanation</h3>
Fluorine F is found in the second column from the right end of a modern periodic table. Fluorine is next to and on the left of the noble gas element neon. A neutral fluorine atom is one electron short of neon, which contains 8 electrons in the outermost shell when neutral. As a result, there are 7 electrons in the outermost shell of a fluorine atom.
Fluorine is in period 2. Its electrons occupy two main shells. The second main shell is the outermost shell of F. There are two subshells in the second main shell:
- 2s, which holds up to two electrons, and
- 2p, which holds up to six electrons.
A 2s electron carries less energy than a 2p electron. By Aufbau principle, the seven electrons will fill the two spaces in 2s before moving on the 2p. Among the 7 outermost shell electrons, 
will end up going to 2p.
The only 2s orbital is filled with two electrons. The two 2s electrons will pair up with opposite spins, as seen with the two arrows. Two of the 2p orbitals will contain two electrons. Those electrons will also pair up. The third 2p orbital will contain only one electron. That electron can spin either 
or 
. Here that electron is shown as an upward arrow.
Answer:
Reflection
Explanation:
Because light is bouncing of the object
Answer:
The correct answer is 4.58 grams.
Explanation:
Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).
For the reaction, n * 96500 C = molar mass
1C = molar mass/n*96500 = Equivalent wt / 96500
w = Equivalent wt / 96500 * I * t
In the given reaction,
Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.
Now putting the values in the equation we get,
w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)
w = 4.58 gm.
