<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.
So the answer is:
8,12,48
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Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
H+= 10^-14 / [OH-1 = 3.125 * 10^-4 M
pH=-log(H+) = 3.505
Just round it down and your answer = 3.5
The charge for this compound is positive. For Fe, it's charge is positive 3, and for OH, it's charge is negative 1. You would then criss cross the charges of each and come out with Fe(OH)3. I hope this helped!! :)
