<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
B) they have no charges and are inside an atom.*
Answer: It will be produced 276,3 mg of product
Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:
C14H10 + C4H2O3 → C18H12O3
The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.
1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.
As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.
Answer: The electronic configuration of Chlorine will be ![[Ne]3s^23p^5](https://tex.z-dn.net/?f=%5BNe%5D3s%5E23p%5E5)
Explanation: Chlorine is an element which belongs to the P-block of the periodic table. Its atomic number is 17.
Atomic Number = Number of electrons
Number of electrons = 17
The nearest noble gas which lies to Chlorine is Neon which has 10 electrons. Rest of the 7 electrons are filled in the 3s and 3p orbitals. Hence, the electronic configuration of Chlorine is
![[Cl]=[Ne]3s^23p^5](https://tex.z-dn.net/?f=%5BCl%5D%3D%5BNe%5D3s%5E23p%5E5)
