F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
Answer:
Ooh sweetie goodluck
Explanation:
May the gods ever be in your favor.
Answer: the answer should and most definitely be D.
Explanation: I mean think about it after a while only a few radioactive nuclei are left which means it will dye down after a while which also makes it very boring hope this helps :)
the mass percent of sugar in this solution is 46%.
Answer:
Solution given:
mass of solute=34.5g
mass of solvent=75g
mass percent=
=
D. due to the the water it will bring sand with the water there for us is D.
