<span> d. Reduced visibility
Acid rain doesn't reduce visibility, smog does
</span>
Answer:
The binding energy released is 1.992 X 10⁻¹⁸ J
Explanation:
Given;
mass of the alpha particle, m = 6.64 x 10⁻²⁷ kg
speed of the alpha particle, c = 3 x 10⁸ m/s
The binding energy released is given by;
![E_b = mc^2](https://tex.z-dn.net/?f=E_b%20%3D%20mc%5E2)
where;
m is mass of the particle
c is speed of the particle
E = 6.64 x 10⁻²⁷ (3 x 10⁸)²
E = 1.992 X 10⁻¹⁸ J
Therefore, the binding energy released is 1.992 X 10⁻¹⁸ J
mass is the answer 110% sure hope this helps
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After 0.0015 mol of NAOH is added to 0.5000 l of this solution, the pH is 3.37.
<h3>Steps</h3>
First, modify the Henderson-Hassle-batch equation to find the acid's pka.
![pH=pka + log\frac{[A]}{[HA]}](https://tex.z-dn.net/?f=pH%3Dpka%20%2B%20log%5Cfrac%7B%5BA%5D%7D%7B%5BHA%5D%7D)
![pka = pH-log\frac{[A]}{[HA]}](https://tex.z-dn.net/?f=pka%20%3D%20pH-log%5Cfrac%7B%5BA%5D%7D%7B%5BHA%5D%7D)
![=3.35- log \frac{[0.1500]}{[ 0.2000 ]}](https://tex.z-dn.net/?f=%3D3.35-%20log%20%5Cfrac%7B%5B0.1500%5D%7D%7B%5B%200.2000%20%5D%7D)
![pka =3.47](https://tex.z-dn.net/?f=pka%20%3D3.47)
Molarity of the added NaOH.
![M=\frac{mol}{L}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7Bmol%7D%7BL%7D)
![=\frac{0.0015 mol}{ 0.5000L}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.0015%20mol%7D%7B%200.5000L%7D)
= 0.003 M
In water, NaOH will dissociate
![NaOH- > Na^{+} + OH^{-}](https://tex.z-dn.net/?f=NaOH-%20%3E%20Na%5E%7B%2B%7D%20%20%2B%20OH%5E%7B-%7D)
strong base OH will react with HA
HA+ OH -->
+ A
the reaction also produced A. as, all OH are consumed in the reaction
HA will be decreased by 0.003 M and A will be increased by 0.003 M.
![HA= 0.2000-0.003 \\= 0.1970](https://tex.z-dn.net/?f=HA%3D%20%200.2000-0.003%20%5C%5C%3D%200.1970)
![A=0.1500+0.003=0.1530 M](https://tex.z-dn.net/?f=A%3D0.1500%2B0.003%3D0.1530%20M)
solving new pH using pka and the new values of [HA] and [A]
![pH=pka + log\frac{[A]}{[HA]}](https://tex.z-dn.net/?f=pH%3Dpka%20%2B%20log%5Cfrac%7B%5BA%5D%7D%7B%5BHA%5D%7D)
![=3.35 + log\frac{[0.1530]}{[0.1970]}](https://tex.z-dn.net/?f=%3D3.35%20%2B%20log%5Cfrac%7B%5B0.1530%5D%7D%7B%5B0.1970%5D%7D)
pH=3.37 is the ph after 0.0015 mol of naoh is added to 0.5000 l of this solution
learn more about buffer here
<u>brainly.com/question/22390063</u>
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