PV = n RT
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin
V= (n RT) / P = 35.8 x 0.082 x 343.15 / 10 = 100.7 ≈ 101 L
V = 101L
The number of particles in one mole is given be Avagadro's number <span>6.022×10^23
Multiply by number of moles.
3 ×10^-21 mol * 6.022 ×10^23 molecules/mol = </span><span>1,807 molecules
(rounded to nearest whole number)
</span>
Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
The answer is (3) HClO. In the Cl2, chlorine has an oxidation number of zero. In HCl, the oxidation number is -1. In HClO2, the oxidation number is +3. In HClO, it is +1. You can calculate this by using O with oxidation number of -2 and H with +1.
