Answer:
3.2 × 10⁻⁸
Explanation:
Let's consider the solution of magnesium carbonate.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The Ksp is:
Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen
The mass of nitrogen in the second sample is 23.33 g.
Explanation: This is a reaction of oxidation of 
in the presence of acidified 
. Acidified is a strong oxidizing agent.
To balance out the 
on the reactant side, we write 
on the product side.
Balancing out the following reaction gives us:
