Answer:Number of electrons that are present in an atom is determined by the electronic configuration of that atom.
If an ion is carrying a positive charge, it means that the atom has lost electrons and if an ion is carrying a negative charge, it means that the atom has gained electrons.
For the given options:
Option A: The atomic number of hydrogen atom is 1 and the electronic configuration for ion will be:
Thus, this atom does not have any electrons.
Option B: The atomic number of bromine atom is 35 and the electronic configuration for ion will be:
Thus, this atom has 36 electrons.
Option C: The atomic number of aluminium atom is 13 and the electronic configuration for ion will be:
Explanation:
Answer:
178.35g
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles ÷ volume
Based on the information provided in this question, molarity (M) of the solution = 1.50 M, volume = 725 mL = 725/1000 = 0.725L, n = ?
1.50 = n / 0.725
n = 1.50 × 0.725
n = 1.0875mol
Molar mass of Na3PO4
23(3) + 31 + 16(4)
= 69 + 31 + 64
= 164g/mol
Mole = mass ÷ molar mass
1.0875 = mass/164
mass = 178.35g
A teratogen is an exposure (substance, organism or process) in pregnancy that has a harmful effect on the fetus. Teratogens can be diseases, medications, drugs and environmental exposures.
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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