The answer would be C. Hope this helps
Explanation:
Fertilizers are chemically synthesized plant nutrients.
Nitrogen (N), Phosphorus (P) and Potassium (K) are macronutrients and are required in large amounts by plants. So, farmers use fertilizers in order to supply these nutrients. NPK 15:15:15 , NPK 20:20:20, NPK 15:30:15 are examples of fertilizers used to supply N, P, K to crops.
Answer:
the atomic packing factor of Sn is 0.24
Explanation:
a = b = 5.83A and c = 3.18A.
Volume of unit cell = a²c
= (5.83)² * 3.18 * 10⁻²⁴ cm³
= 1.08 * 10⁻²²cm³
Volume of atoms =

(∴ BCC, effective number of atom is 2)
Volume of atoms =

= 2.55*10⁻²³cm³


<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
Answer:
While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So,
for the given reaction is -99.4 J/K