To find the molecular formula from the empirical formula, first calculate the empirical formula mass. Then, divide the molecular or molar mass by the empirical formula mass to find the number of empirical formula units per molecule. Finally, multiply the subscript of each element in the empirical formula by this number to get the molecular formula.
The process of finding the molecular formula from the empirical formula involves a few steps. First, you need to determine the empirical formula of a compound, which is the simplest positive integer ratio of atoms present in a compound. After you have the empirical formula, you can calculate the empirical formula mass by summing up the average atomic masses of all the atoms in the empirical formula.
Next, compare the compound's molecular or molar mass (which you should know or have been given) to the empirical formula mass. This comparison is done by dividing the molecular or molar mass by the empirical formula mass. This will give you the number of empirical formula units, denoted as n, per molecule of the compound.
The final step is to take the empirical formula and multiply the subscript of each element in it by n. This will give you the molecular formula of the compound.
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b.0.997 M
c.0.180 M
d.0.156 M
e.55.3 M
Answer:
55.3 M
Explanation:
Given temperature = 25°C
Density = 0.997 g/mL
Based on density the mass of water per L = 997 grams
The molar mass of water = 18
The moles of water =
The molarity is defined as the moles of water per litre.
The moles = 55.3
Volume = 1 L
So molarity = 55.3 /1 = 55.3 M
b. ionic bonding
c. metallic bonding
d. nonpolar covalent bonding
e. polar covalent
Explanation:
An equation that contains equal number of atoms on both reactant and product side is known as a balanced chemical equation.
Whereas when there will be increase in oxidation number of the atom then it is known as oxidation. When there is decrease in oxidation number of the atom then it is known as reduction.
For example,
Oxidation half reaction :
Reduction half reaction :
Therefore, we can conclude that sulfur atom has undergone oxidation and nitrogen atom has undergone reduction.
B) atomic number
C) number of protons
D) number of electrons
The quantity that can vary among atoms of the same element is the mass number, which represents the sum of the number of protons and neutrons in an atom.
The quantity that can vary among atoms of the same element is the mass number. The mass number represents the sum of the number of protons and neutrons in an atom. While the atomic number (which represents the number of protons) remains constant for atoms of the same element, the number of neutrons can vary, resulting in different mass numbers for atoms of the same element.
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(B) 100
(C) 15
(D) 25
(E) 30
MC₂H₄O₂ = 2 * 12 g/mol + 4 * 1 g/mol + 2 * 16 g/mol = 60 g/mol
c = 0.5 mol/L
V = 1 L
n = c * V
n = 0.5 mol/L * 1 L = 0.5 mol
m = n * M
m = 0.5 mol * 60 g/mol = 30 g/mol
answer E
(-_-(-_-)-_-)
The compound Mn3(PO4)2 is correctly named as manganese(II) phosphate, based on the rule of naming ionic compounds and considering the oxidation state of the manganese ion.
The correct name of the compound Mn3(PO4)2 is manganese(II) phosphate. This is based on the rule of naming ionic compounds. With ionic compounds, the metal cation's oxidation state is represented in Roman numerals in parentheses if the metal can have more than one charge state.
In this case, the Mn3+ cation in Mn3(PO4)2 has an oxidation state of +2 since the phosphate ions PO4^3- balance out 3 Mn ions. Therefore, the compound is correctly named manganese(II) phosphate.
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