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:
1.5g/cm³
Explanation:
density=mass÷volume
mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g
volume of the cube = 10×10×10 = 1000cm³
density= divide 1500g÷1000cm = 1.5g/cm³
<h2>
Density= 1.5g/cm³</h2>
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A mutation that gives a rabbit a third ear
Answer:
Substitution mutation
Explanation:
A substitution mutation is a type of mutation in which one or more nucleotide base is replaced by another in a sequence. This will result in the replacement of one or more amino acid in the amino acid sequence.
This is the case in this question where the original amino acid sequence was given as: Leucine – Alanine – Glycine – Leucine. After mutation, the following mutated sequence was produced: Leucine – Alanine – Valine – Leucine.
As illustrated above, one would notice that there is replacement of GLYCINE amino acid by VALINE in the mutated sequence, hence, it is an example of SUBSTITUTION MUTATION.