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 : The Lewis-dot structure of 
is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 1(4) + 2(7) + 1(6) = 24
According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.
Answer:
Radiation is the transfer of heat between objects which are not touching. I think
Explanation:
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
Answer:
Explanation:
Hello!
In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:
Then, the liters of solution:
Finally, the molarity turns out:
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