Answer
The empirical formula is CrO₂Cl₂
Explanation:
Empirical formula is the simplest whole number ratio of an atom present in a compound.
The compound contain, Chromium=33.6%
Chlorine=45.8%
Oxygen=20.6%
And the molar mass of Chromium(Cr)=51.996 g mol.
Chlorine containing molar mass (Cl)= 35.45 g mol.
Oxygen containing molar mass (O)=15.999 g mol.
Step-1
Then,we will get,
Cr=
mol
Cl= 
mol.
O=
mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr=
Cl=
O=
Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
Answer:
139.98 g to nearest hundredth.
Explanation:
Using Avogadro's Number:
One mole (167.26 g) of Erbium equates to 6.022141 * 10^23 atoms.
So 5.04 * 10^23 = 167.26 * 5.04/6.022141
= 139,98 g.
A compound<span> is a </span>pure substance<span> composed of two or more different atoms chemically bonded to one another. A </span>compound<span> can be destroyed by chemical means. It might be broken down into simpler </span>compounds<span>, into its elements or a combination of the two.</span>
Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.<span>
Chemical reaction: HCOOH(aq) + H</span>₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).<span>
c(HCOOH) = 0,1 M.
[</span>H₃O⁺] = [HCOO⁻] = x.<span>
[HCOOH] = 0,1 M - x.
</span>Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).<span>
Solve quadratic equation: x = </span>[H₃O⁺] = 0,0042 M.
