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.
High energy hope this helps
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
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The thing that it is called is it is called the atomic number of the thing
Answer:
the answer to your problem is A.Subscript...BRAINLIEST