The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer:
0.92322835
Explanation:
0.92322835 rounded= 0.9 or 0.92
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I think it’s A sorry if i’m wrong
Answer:
2.5×10^-7mol/dm^3
Explanation:
Firstly convert the cm^3 to dm^3
200×1000=200000dm^3
Calculate the g/dm^3
2/200000=0.00001g/dm^3
To calculate mol/dm^3
Mol/dm^3=mass given\molar mass
=0.00001/40
=2.5×10^-7mol/dm^3