First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
Answer : It increases
Rusting is where oxygen binds to iron and forms iron oxide.
So once iron rusts, there is oxygen, just not in air; it's in the iron oxide.
All reactions are reversible, albeit at different rates (the "irreversible" ones are still reversible, but much slower given that they take so much collision luck and energy.
Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula
Substitution
Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 = 
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm
