A). Carbon enters the atmosphere as carbon dioxide from respiration and combustion.
Answer:
Ag+(aq) + Cl-(aq) <--->AgCl(s)
Ksp = [Ag+] [Cl-]
aii) Pb2+(aq) + 2Cl-(aq) <--->PbCl2(s)
Ksp = [Pb+] [Cl-]^2
Explanation:
In this question, we are to write the net ionic equation and corresponding Ksp expression for the salts in the question.
We proceed as follows;
Ag+(aq) + Cl-(aq) <--->AgCl(s)
Ksp = [Ag+] [Cl-]
aii) Pb+2(aq) + 2Cl-(aq) <--->PbCl2(s)
Ksp = [Pb+] [Cl-]^2
Answer:
3.568 g of H₂O
Solution:
The Balance Chemical Equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
4.04 g (2 mol) H₂ reacts with = 32 g (1 mol) of O₂
So,
0.40 g of H₂ will react with = X g of O₂
Solving for X,
X = (0.40 g × 32 g) ÷ 4.04 g
X = 3.17 g of O₂
It means 0.4 g of H₂ requires 3.17 g of O₂, while we are provided with 3.2 g of O₂ which is in excess. Therefore, H₂ is the limiting reagent and will control the yield of products.
Step 2: Calculate amount of Water produced,
According to equation,
4.04 g (2 mol) of H₂ produces = 36.04 g (2 mol) of H₂O
So,
0.40 g of H₂ will produce = X moles of H₂O
Solving for X,
X = (0.40 g × 36.04 mol) ÷ 4.04 g
X = 3.568 g of H₂O
the configuration for chlorine is [Ne] 3s2 3p5