<span>A full outer shell has 8 valence electrons, and since each nitrogen has only five, they both need three more to get to a full outer shell. However, since they're the same atom, and they need the same amount of electrons, they're going to form a covalent bond, where they both SHARE three (if one gave three to the other, it would be down to two and even MORE unbalanced)</span>
Answer : The [α] for the solution is, -118.8
Explanation :
Enantiomeric excess : It is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.
Mathematically,

Given:
% major enantiomer = 86 %
% minor enantiomer = 14 %
Putting values in above equation, we get:


Now we have to calculate the [α] for the solution.
![[\alpha]=\text{Enantiomer excess}\times [\alpha]_{Pure}](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D%5Ctext%7BEnantiomer%20excess%7D%5Ctimes%20%5B%5Calpha%5D_%7BPure%7D)
![[\alpha]=0.72\times -165](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D0.72%5Ctimes%20-165)
![[\alpha]=-118.8](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D-118.8)
Thus, the [α] for the solution is, -118.8
The answer is (3) 3. The calculation rule of significant figures is when multiplication and division, the significant figures of result is equal to the least of the numbers involved. The significant figure of 2.70 and 80.01 is 3 and 4.
It isn't a materialistic object.
Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 =
Now, 1.687 = square root [
]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol