Answer:
2,5
Explanation:
When move from left to right in periodic table atomic number increases. So element to the right of carbon has one electron more than carbon therefore it's electronic configuration would be 2,5
Answer:
B. The products are nuclei of elements that are different from the original elements.
The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>
Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K

Answer:The formulas of ionic compounds are:
a)
b)
c)
d)
Explanation:
Formulas for the an ionic compounds is determine by:
Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
(a) Copper bromide :Given that it contains
ion.

(b) Manganese oxide : Given that it contains
ion.

(c)Mercury iodide :Given that it contains 

(d) Magnesium phosphate :Given that it contains 

Answer:
C.
will precipitate out first
the percentage of
remaining = 12.86%
Explanation:
Given that:
A solution contains:
![[Ca^{2+}] = 0.0440 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D%200.0440%20%5C%20M)
![[Ag^+] = 0.0940 \ M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%20%3D%200.0940%20%5C%20M)
From the list of options , Let find the dissociation of 

where;
Solubility product constant Ksp of
is 
Thus;
![Ksp = [Ag^+]^3[PO_4^{3-}]](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BAg%5E%2B%5D%5E3%5BPO_4%5E%7B3-%7D%5D)
replacing the known values in order to determine the unknown ; we have :
![8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]](https://tex.z-dn.net/?f=8.89%20%5Ctimes%2010%20%5E%7B-17%7D%20%20%3D%20%280.0940%29%5E3%5BPO_4%5E%7B3-%7D%5D)
![\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]](https://tex.z-dn.net/?f=%5Cdfrac%7B8.89%20%5Ctimes%2010%20%5E%7B-17%7D%7D%7B%280.0940%29%5E3%7D%20%20%3D%20%5BPO_4%5E%7B3-%7D%5D)
![[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D%5Cdfrac%7B8.89%20%5Ctimes%2010%20%5E%7B-17%7D%7D%7B%280.0940%29%5E3%7D)
![[PO_4^{3-}] =1.07 \times 10^{-13}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D1.07%20%5Ctimes%2010%5E%7B-13%7D)
The dissociation of 
The solubility product constant of
is 
The dissociation of
is :

Thus;
![Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=2.07%20%5Ctimes%2010%5E%7B-33%7D%20%3D%20%280.0440%29%5E3%20%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2](https://tex.z-dn.net/?f=%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%280.0440%29%5E3%7D%3D%20%20%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%5E2%20%3D%20%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%280.0440%29%5E3%7D)
![[PO_4^{3-}]^2 = 2.43 \times 10^{-29}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%5E2%20%3D%202.43%20%5Ctimes%2010%5E%7B-29%7D)
![[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D%20%5Csqrt%7B2.43%20%5Ctimes%2010%5E%7B-29%7D)
![[PO_4^{3-}] =4.93 \times 10^{-15}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D4.93%20%5Ctimes%2010%5E%7B-15%7D)
Thus; the phosphate anion needed for precipitation is smaller i.e
in
than in

Therefore:
will precipitate out first
To determine the concentration of
when the second cation starts to precipitate ; we have :
![Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2](https://tex.z-dn.net/?f=2.07%20%5Ctimes%2010%5E%7B-33%7D%20%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%281.07%20%5Ctimes%2010%5E%7B-13%7D%29%5E2)
![[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5E3%20%3D%20%20%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%281.07%20%5Ctimes%2010%5E%7B-13%7D%29%5E2%7D)
![[Ca^{2+}]^3 =1.808 \times 10^{-7}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5E3%20%3D1.808%20%5Ctimes%2010%5E%7B-7%7D)
![[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D%5Csqrt%5B3%5D%7B1.808%20%5Ctimes%2010%5E%7B-7%7D%7D)
![[Ca^{2+}] =0.00566](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D0.00566)
This implies that when the second cation starts to precipitate ; the concentration of
in the solution is 0.00566
Therefore;
the percentage of
remaining = concentration remaining/initial concentration × 100%
the percentage of
remaining = 0.00566/0.0440 × 100%
the percentage of
remaining = 0.1286 × 100%
the percentage of
remaining = 12.86%