find mol of N2 present using gas law equation
PV = nRT
P = pressure = 688/760 = 0.905 atm.
V = 100mL = 0.1L
n = ???
R = 0.082057
T = 565+273 = 838
Substitute:
0.905*0.1 = n*0.082057*838
n = 0.0905 / 68.76
n = 0.00132 mol N2
Molar mass N2 = 28 g/mol
0.00132 mol = 0.00132*28 = 0.037g N2 gas
Answer:
they are made up of hard spheres that are in random motion
Answer:
So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.
Explanation:
What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.
It’s B. Chromium(III) oxide
What are you asking???? If the formula had no atoms of oxygen then......
