Answer:
Kp is 0.228/atm
Explanation:
This is the reaction:
CO + Cl<u>₂</u> → CCl₂O
1 mol of carbon monoxide and 1 mol of chlorine produce 1 mol of phosgene.
Formula for Kp which derivates from Kc is:
Kp = Kc (R.T)ⁿᵇ ⁻ ⁿᵃ
Δп = nb (moles in the products) - nₐ(moles in the reactants)
Δп = 1 - 2 = -1
T is T° in K → T°C + 273 = 611°C +273 = 884K
R → Universal constant gas → 0.082 L.atm/mol.K
We replace the data: Kp = 16.5 L/mol (0.082 . 884K)⁻¹ → 0.228/atm
The empirical formula is Fe₃O₄.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of Fe to O.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio</u>¹ <u>×3</u>² <u>Integers</u>³
Fe 0.77 1 3 3
O 1.0 1.3 3.9 4
¹ To get the molar ratio, you divide each number of moles by the smallest number (0.77).
² If the ratio is not close to an integer, multiply by a number (in this case, 3) to get numbers that are close to integers.
³ Round off these numbers to integers (3 and 4).
The empirical formula is Fe₃O₄.
Molar mass of CeCl3 = 246.475 g/mol
This compound is also known as Cerium(III) Chloride.
Convert grams CeCl3 to moles or moles CeCl3 to grams
Molecular weight calculation:
140.116 + 35.453*3
Given that solubility product of AgCl = 1.8 X 10^-10
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution =

∴ 1.34 x 10^-5 =

∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus,
mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g