Covalent compounds have bonds where electrons are shared between atoms. Due to the sharing of electrons, they exhibit characteristic physical properties that include lower melting points and electrical conductivity compared to ionic compounds.
Answer:
A binary covalent compound is composed of two different elements (usually nonmetals). For example, a molecule of chlorine trifluoride, ClF3 contains 1 atom of chlorine and 3 atoms of fluorine.
Rule 1. The element with the lower group number is written first in the name; the element with the higher group number is written second in the name. Exception: when the compound contains oxygen and a halogen, the name of the halogen is the first word in the name.
Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name.
Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the root of the element name (e.g., fluorine = F, "fluoride" = F-; sulfur = S, "sulfide" = S2-).
Rule 4. Greek prefixes are used to indicate the number of atoms of each element in the chemical formula for the compound. Exception: if the compound contains one atom of the element that is written first in the name, the prefix "mono-" is not used.
Explanation:
First, it is best to know the chemical formula of pyridine which is C5H5N. To determine the number of carbon atoms present in pyridine, multiply 7.05 mol C5H5N with 5 mol C/ 1 mol C5H5N which then results to 35.35 mol of carbon. Then, multiply the answer to Avogadro's number which is 6.022x10^23 atoms. It is then calculated that the number of carbon atoms in 7.05 moles of pyridine is 2.12x10^25 atoms.
Answer:
44 g oxygen are needed.
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of ammonia = 18.2 g
Solution:
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will calculate the number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 18.2 g/ 17 g/mol
Number of moles = 1.1 mol
Now we will compare the moles of ammonia with oxygen from balance chemical equation.
NH₃ : O₂
4 : 5
1.1 : 5/4×1.1 = 1.375 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.375 mol × 32 g/mol
Mass = 44 g
Answer:
The mass of water to be added is 2 pounds
Explanation:
The given parameters are;
The mass of the given solution = 2 pounds
The concentration of the given solution = 30%
The desired concentration of the solution = 15%
The mass, m of the acetic acid in the given solution = 30% × 2 pounds
m = 30/100 × 2 pounds = 0.6 pounds
To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;
15% of X = 0.6 pounds
15/100 × X = 3/20 × 0.6 pounds
∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds
The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds
Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.
