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>
D.) There must be an equal number of atoms of each element on both sides of the equation
Answer:
= 374.90 kPa
Calculation:
As we know atm and kiloPascal are related to each other as,
1 atm = 101.325 kPa
So,
3.70 atm = X
Solving for X,
X = (3.70 atm × 101.325 kPa) ÷ 1 atm
X = 374.90 kPa
Answer:
with the help of the juice contained in it
