Each line represents a single bond (two electrons), and the lone pairs of electrons are shown as dots. Carbon forms single bonds with three hydrogen atoms and one bromine atom, resulting in a stable Lewis structure that satisfies the octet rule for each element and minimizes formal charges.
To draw the Lewis electron-dot diagram for bromomethane (CH3Br), follow these steps:
Determine the total number of valence electrons for each element in the molecule:
Carbon (C) has 4 valence electrons.
Hydrogen (H) has 1 valence electron.
Bromine (Br) has 7 valence electrons.
Calculate the total number of valence electrons by adding the contributions from each element:
Carbon: 4 electrons x 1 atom = 4 electrons.
Hydrogen: 1 electron x 3 atoms = 3 electrons.
Bromine: 7 electrons x 1 atom = 7 electrons.
Total = 4 + 3 + 7 = 14 electrons.
Determine the central atom. In CH3Br, carbon is the central atom.
Connect the central carbon atom to the surrounding hydrogen and bromine atoms using single bonds (each bond consists of two electrons).
Distribute the remaining valence electrons as lone pairs on the outer atoms to satisfy the octet rule. Hydrogen can only accommodate two electrons (a duet), and bromine can expand its octet.
Calculate the formal charges to ensure that the Lewis structure represents the most stable arrangement of electrons. In CH3Br, the formal charges should ideally be as close to zero as possible.
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Answer:
Ionic bond is formed by the cations and anions and their attractive forces.
B. the direction of its orbit
C. the stability of its orbit
consuming energy
exothermic
endothermic
B. What substances make up an eggshell?
C. Which birds have the most beautiful feathers?
D. Which type of bird makes the best pet in an apartment?
After reading a book about parrots, Tani wants to learn more about them. The question that could be answered through scientific investigation is letter B, ‘What substances make up an eggshell?’ The other choices cannot be answered through scientific investigation.
Answer:
B
Explanation:
Answer:
To calculate the volume of the approximately 0.014 M copper (II) sulfate solution needed to prepare 100.00 milliliters of a 0.00028 M copper (II) sulfate solution, you can use the formula:
(Volume 1)(Concentration 1) = (Volume 2)(Concentration 2)
So, rearranging the formula, you can solve for Volume 1 (the volume of the 0.014 M solution):
Volume 1 = (Volume 2)(Concentration 2) / Concentration 1
Plugging in the values:
Volume 1 = (100.00 milliliters)(0.00028 M) / 0.014 M
Volume 1 ≈ 2.00 milliliters
Approximately 2.00 milliliters of the approximately 0.014 M copper (II) sulfate solution are needed to prepare 100.00 milliliters of a 0.00028 M copper (II) sulfate solution.