Answer: A
0 degrees and 101 kPa are the conditions that describe the standard temperature and pressure. When expressed in K, the standard temperature 0 degrees equals 273.5 K. Also the standard pressure 101 kPa equals 760 mmHg or 1 Atm.
Answer:
56°
Explanation:
First calculate ![a:](https://tex.z-dn.net/?f=a%3A)
![a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}](https://tex.z-dn.net/?f=a%3D2%20R%20%5Csqrt%7B2%7D%3D2%280.1246%29%20%5Csqrt%7B2%7D%3D0.352%20%5Cmathrm%7Bnm%7D)
The interplanar spacing can be calculated from:
![d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}](https://tex.z-dn.net/?f=d_%7B111%7D%3D%5Cfrac%7Ba%7D%7B%5Csqrt%7B1%5E%7B2%7D%2B1%5E%7B2%7D%2B1%5E%7B2%7D%7D%7D%3D%5Cfrac%7B0.352%7D%7B%5Csqrt%7B3%7D%7D%3D0.203%20%5Cmathrm%7Bnm%7D)
The diffraction angle is determined from:
![\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D%5Cfrac%7Bn%20%5Clambda%7D%7B2%20d_%7B111%7D%7D%3D%5Cfrac%7B1%280.1927%29%7D%7B2%280.2035%29%7D%3D0.476)
Solve for ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![\theta=\sin ^{-1}(0.476)=28^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Csin%20%5E%7B-1%7D%280.476%29%3D28%5E%7B%5Ccirc%7D)
The diffraction angle is:
![2 \theta=2\left(28^{\circ}\right)=56^{\circ}](https://tex.z-dn.net/?f=2%20%5Ctheta%3D2%5Cleft%2828%5E%7B%5Ccirc%7D%5Cright%29%3D56%5E%7B%5Ccirc%7D)
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Option-C: They react mainly by substitution.
Explanation:
Alkene doesn't give substitution reactions because they are non polar in nature. The double bond in alkene is responsible for Electrophillic Addition reactions as it electron rich and nucleophilic in nature. Reaction of Alkene is given below,</span>