Answer:
i. Keq=4157.99.
ii. More hydrogen sulfide will be produced.
Explanation:
Hello,
i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:
![Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5BS_2%5D%7D%20%3D%5Cfrac%7B%280.97M%29%5E2%7D%7B%280.051M%29%5E2%280.087%29%7D%20%3D4157.99)
ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.
Best regards.
The colour of copper sulphate solution is blue
Answer:
One: <u>Selenium</u> is Paramagnetic
Explanation:
Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.
Electronic configuration of Scandium;
Sc = 21 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹
Sc³⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶
Hence in Sc³⁺ there is no unpaired electron.
Electronic configuration of Bromine;
Br = 35 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵
Br⁻ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶
Hence in Br⁻ there is no unpaired electron.
Electronic configuration of Magnesium;
Mg = 12 = 1s², 2s², 2p⁶, 3s²
Mg²⁺ = 1s², 2s², 2p⁶
Hence in Mg²⁺ there is no unpaired electron.
Electronic configuration of selenium;
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
Or,
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹
Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
Answer:
Because it only needs one more electron to get to a full valence shell (8), so it really wants it and is pulling other electrons in. It also has to do with needing one more electron to fill the 2p shell. It is a small element which means its electrons are pulled tightly to the nucleus.
Hope this helps!
Explanation:
