Mass CoCl2 = 10.27 g
moles CoCl2 = 10.27 g/ 129.839 g/mol=0.07910
mass water = 17.40 - 10.27=7.13 g
moles water = 7.13 / 18.02 g/mol=0.396
0.396/ 0.07910=5
CoCl2 * 5 H2O
moles CaF2 = 85.8 g/ 78.0748 g/mol=1.10
moles Ca = 1.10
mass Ca = 1.10 x 40.078 g/mol=44.1 g
V = 44.1 / 1.55 =28.5 mL
Greetings!!
21 kg x [(3 x 35.45)/(12.01 + 19.00 + (3 x 35.45))] =
21 kg x (106.35/137.36) = 16.3 kg of chlorine
You just multiply the weight of the material by the fraction of chlorine (by weight). The others are done the same way
Answer:
23.34 %.
Explanation:
- The percentage of water must be calculated as a mass percent.
- We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
- <em>Atomic masses of the elements:</em>
Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.
- <em>Mass of the formula unit:</em>
Cd(NO₃)₂•4H₂O
mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.
- <em> Mass of water in the formula unit:</em>
<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.
- <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
<span> the first ionization </span>energy<span> of an element is the </span>energy<span> needed to</span>remove<span> the outermost, or highest </span>energy<span>, </span>electron<span> from a neutral </span>atom<span> in the gas phase.</span>