The reaction is provided in the attachment and it showing the product 4-methoxy-3-tertiarybutylanisole. In the product, first methoxy group is ortho to tertiary butyl group whereas the second methoxy is present in meta position.
The reactant 4-methoxyanisole is ortho and para directing group whereas both the para position are occupied by both the methoxy group in the ring and thus only ortho position is left with respect to each methoxy and having high electron density and therefore, the coming tertiary butyl group attach to any of the ortho position to methoxy to produce 4-methoxyanisole .
6:12:6: - this simplifies to 1:2:1 (answer)
You get the simplest form by dividing each of the 3 numbers by 6.
The nucleus is the part of the atom where energy is released after said atom is split.
Hope this helped :3
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
Graphing is helpful because it gives us B. A visual representation of the data