The hydrogen and oxygen<span> atoms from H</span>₂O are <span>bonded together through covalent </span>bonding.
Answer:
![[H^+]=1.78x10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.78x10%5E%7B-8%7DM)
Explanation:
Hello there!
In this case, according to the given information about the pH, it is firstly necessary for us to remember that the pH is defined as the potential of the hydrogen ions in the solution and the concentration of those ions represents how many of them are present in the solution; in such a way, it is possible for us use:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Whereas the concentration of hydrogen ions can be calculated as follows:
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
So we plug in the given pH to obtain:
![[H^+]=10^{-7.75}=1.78x10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-7.75%7D%3D1.78x10%5E%7B-8%7DM)
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Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,

When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation,
= 102 pm
Radius of cation,
= 196 pm

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
Answer:
c
Explanation:
esuba is activation energy