I think the next full moon July 12th I think.. I'm pretty sure.
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
<em><u>Answer:</u></em>
<h3>look at the balanced chemical equation for this synthesis reaction
</h3><h3> S mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.</h3><h3 />
Answer:
[He]: 2s² 2p⁵.
[Ne]: 3s².
[Ar]: 4s² 3d¹⁰ 4p².
[Kr]: 5s² 4d¹⁰ 5p⁵.
[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².
Explanation:
- Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.
He contains 2 electrons fill 1s (1s²).
So, [He] can be written before the electronic configuration of 2s² 2p⁵.
Ne contains 10 electrons fill (1s² 2s² 2p⁶).
So, [Ne] can be written before the electronic configuration of 3s².
Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).
So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².
Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).
So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.
Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).
So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².
