Answer:
1335.12 mL of H2O
Explanation:
To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).
From the molarity formula we can first calculate the volume of the solution:
The volume of the solution as we said previously is:
Solution volume = solute volume + solvent volume
To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:
Now with the density of NaOH the milliliters of solute can be determined:
Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:
Solvent volume = solution volume - solute volume
Solvent volume = 1429 mL - 93.88 mL = 1335.12 mL of H2O
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Answer:
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Hello!
After the addition of a small amount of acid, a reasonable value of buffer pH would be
5,15.
If initially there are equal amounts of a weak acid and its conjugate base, the pH would be equal to the pKa, according to the
Henderson-Hasselbach equation:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} ) \\ \\ if [A^{-}]=[HA] \\ \\ pH=pKa + log (1) \\ \\ pH=pKa + 0=5,25](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29%20%5C%5C%20%5C%5C%20if%20%5BA%5E%7B-%7D%5D%3D%5BHA%5D%20%5C%5C%20%5C%5C%20pH%3DpKa%20%2B%20log%20%281%29%20%5C%5C%20%5C%5C%20pH%3DpKa%20%2B%200%3D5%2C25)
So, when adding a little amount of acid the pH should be only a little lower than the pKa. The value from the list that is a little lower than the pKa is
5.15
Have a nice day!
Answer:
Explanation:
Electron affinity is the energy released in adding an electron to a neutral atom in the gas phase.
It is a measure of the readiness of an atom to gain an electron. This property is very peculiar to non-metals. The higher the value, the greater the tendency to accept electrons.
Across a period electron affinity increases due to the increasing nuclear charge not being compensated for.
Down a group, electron affinity decreases due to the low nuclear charge and the large atomic radii.
The exception to this rule is the stability of half-filled sublevels. For example, nitrogen has a configuration of 2,5 with sublevel notation of 1s²2s²2p³.
The p-sublevel has a degeneracy of three and the three electrons goes in singly. This makes the configuration stable.
We expect such an atom to have a higher electron affinity but its configuration is stable and carbon would have a higher affinity than it across the same period.
Half filled sublevels are exception to the trend of electron affinity.
