For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
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Answer:
(2) Organelles must work together and their
activities must be coordinated
Explanation:
Organelles are usually located in cells. They are saddled with the role of performing specific functions in the cells for the overall functioning of life. In eukaryotic cells, the organelles are membrane bounded but in prokaryotic or primitive cells such is not the case.
Examples of cell organelles are ribosome, food vacuole, nucleus e.t.c. Just like organs in the body, organelles must work together in order to enhance life.
Answer:
This question is incomplete
Explanation:
The question is incomplete because of the absence of the table but since the question says there are data from an investigation about a plant growth and five other plants (making six) of the same type, the best way to display this type of data for analyst is to use the grouped bar chart. <u>The grouped bar chart will display the data obtained (from an investigation on plant growth) from different students on each of the six plants (of the the same type)</u>.
Colours are usually used to identify the bars (of a group) or could be used to separate the group from other groups but in this case, colours are better used to identify the bars of a group.
Explanation:
Percentage composition of oxygen = (80/134) * 100% = 59.7%.