Carbon:
1s is filled. 2s is filled. 2p is shown to contain two electrons in one orbital and no electrons in the other two orbitals.
Answer:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.03901 mol/liter
- [Cl₂O] = 0.02351 mol/liter
Explanation:
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<u>1. Chemical reaction:</u>
![H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)](https://tex.z-dn.net/?f=H_2O%28g%29%2BCl_2O%28g%29%5Crightleftharpoons%202%20HOCl%28g%29)
<u>2. Initial concentrations:</u>
i) 1.3 g H₂O
- Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
- Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter
ii) 2.2 g Cl₂O
- Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
- Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter
<u>3. ICE (Initial, Change, Equilibrium) table</u>
![H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)](https://tex.z-dn.net/?f=H_2O%28g%29%2BCl_2O%28g%29%5Crightleftharpoons%202%20HOCl%28g%29)
I 0.0481 0.0326 0
C -x -x +x
E 0.0481-x 0.0326-x x
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<u>4. Equilibrium expression</u>
![K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cdfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D.%5BCl_2O%5D%7D)
![0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}](https://tex.z-dn.net/?f=0.09%3D%5Cdfrac%7Bx%5E2%7D%7B%280.0481-x%29%280.0326-x%29%7D)
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<u>5. Solve:</u>
![x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0](https://tex.z-dn.net/?f=x%5E2%3D0.09%28x-0.0481%29%28x-0.0326%29%5C%5C%5C%5C0.91x%5E2%2B0.007263x-0.000141125%3D0)
Use the quadatic formula:
![x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-0.007263%5Cpm%20%5Csqrt%7B%280.007263%29%5E2-4%280.91%29%28-0.000141125%29%7D%7D%7B2%280.91%29%7D)
The positive result is x = 0.00909
Thus the concentrations are:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
- [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter
They are metric units of measurement.
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So we can conclude that communication skills were developed so knowledge could be passed along to younger generations.
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