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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Hey there mate ;), Im Benjemin and lets solve your question.
★ (Alkanes) : forms single bonds between carbon atoms.
The first four elements are gases and others are liquid in state.
★(Alkenes) : forms double bonds between carbon atoms.
The first three alkenes are gases and rest are liquid.
★ (Alkynes) : forms triple bonds between carbon atoms.
First three are gases and the last one is liquid.
According to boiling point :
The larger structure of the hydrocarbons, the higher the boiling points they have.
In the 3 tables, we can see that the boiling point increases.
Answer:
<h2>The answer is 334 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of ethanol = 423 cm³
density = 0.789 g/cm³
So we have
mass = 0.789 × 423 = 333.747
We have the final answer as
<h3>334 g</h3>
Hope this helps you
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
