Answer is: 5.22·10²² atoms of Iodine.
m(CaI₂) = 12.75 g; mass of calcium iodide.
M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.
n(CaI₂) = m(CaI₂) ÷ M(CaI₂).
n(CaI₂) = 12.75 g ÷ 293.9 g/mol.
n(CaI₂) = 0.043 mol; amount of calcium iodide.
In one molecule of calcium iodide, there are two iodine atoms
n(I) = 2 · n(CaI₂).
n(I) = 0.086 mol; amount of iodine atoms.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(I) = n(I) · Na.
N(I) = 0.086 mol · 6.022·10²³ 1/mol.
N(I) = 5.22·10²²; number of iodine atoms.
Answer:
See below
Explanation:
Molecular formula ( just write down all of the elements ) C 4 H4 O4
Empiracle formual CHO
"Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound"
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows:
Mg=24 g/mole
Cl=38 g/mole
Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so:
(38 g/mole*2 moles)+24 g/mole=100 g/mole
Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows:
3.00 moles * 100 g/mole = 300 g</span>
