Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.
F. because electronegativity generally increases as you move from left to right across a periodic table, and F is farther right than O
Hello!
The correct answer is 1. KCI.
I really hope this helped you out! c:
Using Avogadros number, we can get that 1 mole of an atom
contain 6.022 x 10^23 atoms. Therefore we can use this conversion factor to get
the number of moles:
moles ZnCO3 = 6.11 x 10^22 atoms * (1 mole / 6.022 x 10^23
atoms) = 0.10146 moles
The molar mass of ZnCO3 is about 125.39 g/mol, therefore the
mass is:
mass ZnCO3 = 0.10146 moles * (125.39 g / mol)
<span>mass ZnCO3 = 12.72 g</span>
