Answer:
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
Explanation:
There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.
One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.
Density of a solution is mass of solution per unit volume
Density = mass/volume
mass of solution is 46.08 g
volume of solution is 58.9 mL
since mass and volume is known, density can be calculated
density = 46.08 g / 58.9 mL = 0.78 g/mL
Answer :
(a) The number of moles of D produced can be, 6.67 moles.
(b) The volume of D prepared can be, 24.5 L
Explanation :
The given chemical reaction is:
Part (a) :
From the balanced chemical reaction, we conclude that:
As, 3 moles of A react to give 5 moles of D
So, 4 moles of A react to give 
moles of D
Thus, the number of moles of D produced can be, 6.67 moles.
Part (b) :
As we know that 1 moles of substance occupies 22.4 L volume of gas.
As, 
volume of B gives 
volume of D
As, 9.8 L volume of B gives 
volume of D
Thus, the volume of D prepared can be, 24.5 L
<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M
<u>Explanation:</u>
We are given:
Initial concentration of HI = 1.0 M
The given chemical equation follows:
<u>Initial:</u> 1.0
<u>At eqllm:</u> 1.0-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.037 M
Hence, the concentration of hydrogen gas at equilibrium is 0.037 M
