At the 2nd energy level, in main groups, the 2s orbital is always fully occupied before the 2p orbitals are occupied?!
I have an electrical circuit consisting of a coil and a resistor. I've given it all the possible values (inductance, voltage in the circuit, current in the circuit, voltage across the resistor, ohmic resistance, coil resistance, total impedance, etc.). How can I determine and justify whether the current leads the voltage by PI/2 or PI/4?
Hey guys, I'm a bit stuck with my Excel spreadsheet… I have a “Function List” workbook in which I work and a “Database” workbook in which all components with corresponding data points are listed. Now, the "Function List" folder allows me to select all the components I need from the "Database" folder using a dropdown…
Non-alcoholic beverages can contain up to 0.5% alcohol. At what liter level would non-alcoholic beer have an impact on the measurable blood alcohol content and be detectable on breath alcohol? Or is it even realistically possible to detect alcohol-free beer (0.5%) in breath alcohol?
An old house burned down near me. How long does the smoke linger after a fire? I'd be interested to know.
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Hey, that’s a very good question!
It is not possible to say that water cleaves sodium chloride, but it dissolves the salt, i.e. it dissolves the binding of the salt ions. Salts have an ionic bond, which means that a part (sodium) is positively charged and the other (chloride) is negatively charged. These charges attract strongly and neutralize each other, which is why salts are so stable.
Water is now a very polar molecule. The oxygen atom removes electrons from the H atoms and is thus negatively polarized, while the H atoms are positively polarized. The water molecules are now also attracted by the charges of the salt and arrange around the salt ions; the electronegative O is attracted by the Na+ and the electropositive H by the Cl-. Thus, the water atoms formally push between the binding of Na and Cl and neutralize their charges a bit far. Their attraction therefore decreases more and more until the binding completely breaks down and the salt ions individually relocated by water molecules float in the solution. They are now released and the energy released is called “hydration energy”.
Ah! I miss chemistry at school:(My Abi was not long ago, but chemistry was always my favorite subject:(
You can go to study chemistry, I did it for the reason and it is worth it;)
Ahh okay then becomes difficult, but that also sounds very exciting! 🙂
I already study product design 😀
No water does not cleave the saline but does it leak. It is furthermore sodium chloride to split it you would have to direct current (Gleischstrom) through the salt solution then the NaCl splits into its two elements.
Water NaCl can dissolve because water is nonpolar and NaCl polar.
Thank you. That was very helpful. Is this the case with ammonium chloride? The equation is yes: NaOh + H2O —-> Na + OH + H2O
That’ll be solved or not?
Yes, that’s true, but the ions are still loaded, so Na+ and OH. It is exactly the same principle as I have formulated in my answer 🙂
Ammonium chloride is not “NaOH”, that would be sodium hydroxide. Ammonium chloride is “NH4Cl” and is split into NH4+ and Cl.
Water is not at all nonpolar, on the contrary, water is the most polar solvent at all and therefore it can dissolve salts at all.
Yes, of course. I clearly have a cup of coffee too little. I’ll go to the corner to be ashamed.
But it will be dissolved. However, OH is a charged molecule and not a charged atom. Water cannot split up molecular bonds, because those are covalent. Water can only cleave ionic bonds. That’s why OH remains together. However, the salt NaOH was dissolved, the ionic bond between Na+ and OH no longer exists. :
Yes I forgot but I notice when sodium hydroxide reacts with water it looks like NaOH + H2O—>
Na (+)+OH(-)+H2O and this is sodium hydroxide solution but it is not dissolved although this is actually the same principle as in the other two examples?
You know, enjoy your coffee ^