Answer:
[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹
Explanation:
The balanced equation is
H₂O + Cl₂O ⇌ 2HOCl
Data:
Kc = 0.0900
[H₂O] = 0.004 32 mol·L⁻¹
[Cl₂O] = 0.004 32 mol
1. Set up an ICE table.
2. Calculate the equilibrium concentrations
![K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHOCl%5D%24%5E%7B2%7D%24%7D%7D%7B%5Ctext%7B%5BH%24_%7B2%7D%24O%5D%5BCl%24_2%24O%5D%7D%7D%20%3D%20%5Cdfrac%7B%282x%29%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%3D%200.0900%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B4x%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%26%3D%26%200.0900%5C%5C%20%5Cdfrac%7B2x%20%7D%7B0.00432%20-%20x%7D%20%26%20%3D%20%26%200.300%5C%5C2x%20%26%20%3D%20%26%200.300%280.00432%20-%20x%29%5C%5C2x%20%26%20%3D%20%26%200.001296%20-%200.300x%5C%5C2.300x%20%26%20%3D%20%26%200.001296%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B5.63%5Ctimes%2010%5E%7B-4%7D%7D%5C%5C%5Cend%7Barray%7D)
[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹
[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹
Check:
OK.
The characteristics of mixtures are:
a) they are made up of two of more substances
b) the two substances can be separated by physical means
c) there are two kinds of mixtures : homogenous and heterogenous
d) homogenous mixtures do not have any physicaly distinguished boundary
Answer:
1/ Using the first letter of the name of the elements.
2/ If the name of two or more elements begins with the same letter,the second letter of their name is also taken with first letter.
3/ Symbols are made by using the letters of the name of elements in other languages also.
The element of lowest atomic number whose electron configuration has 4 completely filled p orbitals is, Xenon, Xe with electron configuration:1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶.
According to the question, we are required to identify the element with the lowest atomic number whose electronic configuration has four completely filled p sub-shells.
Evidently, the shell with energy level one has no p orbital.
- Therefore, the existence of p orbitals in electron configuration begins with shell energy level, n=2.
- However, we must know that the p-orbital comprises of 3 sub-orbitals namely;
p(x) , p(y) and p(z) with two electrons of opposing spin in each sub orbital.
- Therefore, the total number of electrons in the p orbital is 6 electrons.
In essence, the element in question should have electron configuration;
- 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶.
The electron configuration above contains 54 electrons and has atomic number, 54.
In essence, the element of lowest atomic number whose electron configuration has 4 completely filled p orbitals is, Xenon, Xe.
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