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:
613 mg
Explanation:

Number of fargday's 
Here, I = 9.20 A
t = 10.5 min
= 10.5 x 60 seconds
So, 

= 0.0208 F
Here, 2e, 2F
2F = 1 mol of Ni

1 mol = 59 gm of Ni
0.0104 mol = 59 x0.0104 gm Ni
= 0.613 gm Ni
= (0.613 x 1000 ) mg of Ni
= 613 mg of Ni
Hi Sara
I would say it is an amorphous solid.
Because it's gradually loses shape.
I hope that's help:)
Answer :
The number of bonding pairs of electrons around the hydrogen atom = 2
The number of lone pairs of electrons around the hydrogen atom = 0
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electrons and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in
= 1 + 4 + 5 = 10
According to Lewis-dot structure we conclude that, there are 8 number of bonding electrons and 2 number of non-bonding electrons.
The number of bonding pairs of electrons around the hydrogen atom = 2
The number of lone pairs of electrons around the hydrogen atom = 0