Answer:
2-ethoxy-2-methylpropan-1-ol
Explanation:
On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid ). The acid medium will produce the <u>hydronium ion</u> (). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.
See figure 1
I hope it helps!
In cell biology, the cytoplasm is the material or protoplasm within a living cell, excluding the cell nucleus. It comprises cytosol (the gel-like substance enclosed within the cell membrane) and the organelles – the cell's internal sub-structures. All of the contents of the cells of prokaryote organisms (such as bacteria, which lack a cell nucleus) are contained within the cytoplasm. Within the cells of eukaryote organisms the contents of the cell nucleus are separated from the cytoplasm, and are then called thenucleoplasm. The cytoplasm is about 80% water and usually colorless.[1]
It is within the cytoplasm that most cellular activities occur, such as many metabolic pathways including glycolysis, and processes such as cell division. The concentrated inner area is called the endoplasm and the outer layer is called the cell cortex or theectoplasm.
Movement of calcium ions in and out of the cytoplasm is a signaling activity for metabolic processes.[2]
In plants, movement of the cytoplasm around vacuoles is known as cytoplasmic streaming.
Explanation:
from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x
6.0*2 moles/ 1 mole
= 12 moles
this implies that, 6.0 moles of O2 will give = 12 moles of water
<u>Answer:</u> The mass of water that should be added in 203.07 grams
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
m = molality of barium iodide solution = 0.175 m
= Given mass of solute (barium iodide) = 13.9 g
= Molar mass of solute (barium iodide) = 391.14 g/mol
= Mass of solvent (water) = ? g
Putting values in above equation, we get:
Hence, the mass of water that should be added in 203.07 grams