<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em><u>Proton</u></em><em><u>.</u></em>
<em><u>Additional</u></em><em><u> </u></em><em><u>Information</u></em><em><u>:</u></em>
<em><u>There</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>three</u></em><em><u> </u></em><em><u>types</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>subatomic</u></em><em><u> </u></em><em><u>particles</u></em><em><u>.</u></em><em><u> </u></em><em><u>They</u></em><em><u> </u></em><em><u>are</u></em><em><u>:</u></em>
- <em><u>Proton</u></em>
- <em><u>Electron</u></em>
- <em><u>Neutron</u></em>
<em><u>Proton</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>positively</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u>,</u></em><em><u>Electron</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>negatively</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>Neutron</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>charge</u></em><em><u> </u></em><em><u>less</u></em><em><u>.</u></em>
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>be</u></em><em><u> </u></em><em><u>helpful</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>you</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
0.84545454545 g/ml
Explanation:
to find density do mass/volume
When two waves come in contact with each other while traveling on the same medium, its called constructive interference. Constructive interference happens when two waves come together.
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%5D_%7Bdiluted%7D%7D%7B%5BCl%5E%7B-%7D%5D_%7Bconcentrated%7D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V