The correct answer is option C, 5.02 x 10²² carbon atoms
Atomic mass of C = 12 g/mol
According to Avogadro, 1 mole of C has 6.023 x 10²³C atoms
Now 1 mole of C is equal to 12 g
Therefore, 12 g of C = 6.023 x 10²³ C atoms
1 g of C =
C atoms = 5.02 x 10²² C atoms
valance electrons that reside in the outermost shell of an atom in the highest energy level. They are important to atoms because the fewer valence electrons that the atom holds, it becomes less stable.
I take honors chemistry I hope this helps.
Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Explanation:
Integrated rate law for second order kinetics is given by:
k = rate constant = 
= initial concentration = 0.860 M
a= concentration left after time t = 0.260 M

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Answer:
the mass of the air in the room is 4.96512 kg ( in 0°C)
Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2