<em>Calculate the pH of the following substances formed during a volcanic eruption:
</em>
<em>• Acid rain if the [H +] is 1.9 x 10-5
</em>
<em>• Sulfurous acid if [H +] = 0.10
</em>
<em>• Nitric acid if [H +] = 0.11</em>
<em />
<h3>Further explanation </h3>
pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.
pH = - log [H⁺]
![\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1.9%5Ctimes%2010%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D5-log1.9%5C%5C%5C%5CpH%3D4.72)
![\tt pH=-log[10^{-1}]\\\\pH=1](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B10%5E%7B-1%7D%5D%5C%5C%5C%5CpH%3D1)
![\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B11%5Ctimes%2010%5E%7B-2%7D%5D%5C%5C%5C%5CpH%3D2-log~11%3D0.959)
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
Nitrogen (around 78%), Oxygen (around 21%), and Argon (around 1%).
Hope this helps :)
To Earth, since it has the same radio and masses. This is what I believe.
