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tigry1 [53]
2 years ago
5

How much energy would be released as an electron moved from the n=4 to the n=3 energy level?

Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
8 0

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

                                     E_n =  -13.6/n^{2}  eV

where n is the energy level

Therefore,

                   E_4 = -13.6/4^{2} \\E_3  = -13.6/3^{2}

Thus, E_4  =   -0.85eV

         E_3  =  -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

                                       E_4 -E_3

                                   =  -0.85 - ( -1.51)

                                   = 0.66eV

To know more about energy released in electron transition

brainly.com/question/8384785

#SPJ4

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How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium
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0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as \text{Pb}^{2+} ions. The formula for a nitrate ion is {\text{NO}_3}^{-}. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each \text{Pb}^{2+} ion in lead nitrate would pair up with two {\text{NO}_3}^{-} ions. The formula for lead nitrate will be \text{Pb}({\text{NO}_3})_2. Each formula unit of lead nitrate will contain one \text{Pb}^{2+} ion and two {\text{NO}_3}^{-} ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as \text{Cr}^{3+}. Similarly, each \text{Cr}^{3+} will pair up with three {\text{NO}_3}^{-} ions. The formula for chromium (III) nitrate will be \text{Cr}(\text{NO}_3})_3. Each formula unit of chromium (III) nitrate will contain one {\text{NO}_3}^{-} ion and three {\text{NO}_3}^{-} ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of {\text{NO}_3}^{-} ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

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