<h3><u>Answer</u>;</h3>
Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.
<h3><u>Explanation;</u></h3>
- Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
- Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
- <em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
- <em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>
Answer is: A. 1.81 mol.
Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.
n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.
n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.
From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.
n(Fe(OH)₂) = n(KOH) ÷ 2.
n(Fe(OH)₂) = 3.62 mol ÷ 2.
n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.
Answer:
Water
Explanation:
Water is universal solvent it can disolves other things
Answer:
Equilibrium constant for 
is 0.5
Equilibrium constant for decomposition of 
is 
Explanation:
dissociates as follows:
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for is 0.5
Now calculate the equilibrium constant for decomposition of
It is given that 
is decomposed.
decomposes as follows:
initial 1.0 M 0 0
at eq. concentration of is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
![[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D3.3%5Ctimes%2010%5E%7B-5%7D%20%3D%203.3%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of is
