Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
Answer:
An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. In general, the size of an atom will decrease as you move from left to the right of a certain period.
Explanation:
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
An ionized has consisting of positive ions and free elections in proportion resulting in more or less no overall electric charge
The atomic mass would be 28.08535 amu. Multiply 27.9769 by .92297 = 25.803. Multiply 28.9765 by .046832 to get 1.357. Multiply 29.9738 by .03872 to get .925351136. Add 25.803 + 1.357 + .03872 to get 28.08535 amu
