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Triss [41]
4 years ago
4

To show the electron configuration for an atom, what is the advantage of using an orbital notation compared to a dot structure?

Orbital notation takes up less space. Orbital notation shows the spin of the electrons. Orbital notation makes it easier to count the electrons. Orbital notation shows the electron distribution in shells.
Chemistry
2 answers:
RideAnS [48]4 years ago
8 0
Correct answer: Option B i.e. <span>Orbital notation shows the spin of the electrons.

Reason: 
Dot diagram provides only the electrons that are present in valance shell. Furthermore, it provides no information about the electron spin. On other hand, </span>orbital notation is writing taking into consideration aufbau principle, Hund's rule and Pauli's exclusion principle. Hence, it provides correct description of electron distribution of electrons in atoms. Furthermore, <span>orbital notation shows the spin of the electrons. </span>
LiRa [457]4 years ago
4 0

<u>Answer:</u> The correct answer is Orbital notation shows the spin of the electrons.

<u>Explanation:</u>

Advantages of Orbital notation :

  • One can easily count total numbers of electron in an atom
  • It conveys the distribution of electrons in the shells.
  • It also tells us about the spin of the electrons.

Where as spin of the electrons is not conveyed by the electron dot structure.

Hence, the corrects answer is : Orbital notation shows the spin of the electrons.

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AnnZ [28]

Answer:

\frac{\textup{2}}{h}

Explanation:

Given:

Depth = h

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also,

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= \frac{\rho gh\times w\times1}{\frac{1}{2}\times\rho gh\times(h\timesw)}

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Sav [38]

Explanation:

(6.1).    The reaction equation will be as follows.

             BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)

Assuming the value of K_{sp} as 1.1 \times 10^{-10} and let the solubility of each specie involved in this reaction is "s". The expression for K_{sp} will be as follows.

            K_{sp} = [Ba^{2+}][SO^{-}_{2}]    (Solids are nor considered)

                        = s \times s

                   s = \sqrt{K_{sp}}

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Therefore, solubility of barium sulfate in water is 1.05 \times 10^{-5}.

(6.2).   As the molar mass of BaSO_{4} is 233.38 g/mol

Therefore, the solubility is g/L will be calculated as follows.

                233.38 g/mol \times 1.05 \times 10^{-5}

                  = 2.45 \times 10^{-3} g/L

Therefore, solubility of barium sulfate in grams per liter is 2.45 \times 10^{-3} g/L.

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