The picture of Au₃N is attached below.
The first part of the picture shows the formation of Au and N ions.
Formation of Au⁺¹ :
As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.
Formation of N⁻³ :
As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)
Formation of Au₃N:
Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.
The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.
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<h2>C</h2>
Explanation:
The atomic number of S is 
So,number of electrons in S is 
The electronic configuration of S is 
The orbital electronic configuration of S is 
So,the number of orbitals involved is 9.
The frequency stays the same it just gets louder
Answer:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L