pH solution = 8.89
<h3>Further explanation</h3>
Given
The concentration of HBr solution = 1.3 x 10⁻⁹ M
Required
the pH
Solution
HBr = strong acid
General formula for strong acid :
[H⁺]= a . M
a = amount of H⁺
M = molarity of solution
HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1
Input the value :
[H⁺] = 1 x 1.3 x 10⁻⁹
[H⁺] = 1.3 x 10⁻⁹
pH = - log [H⁺]
pH = 9 - log 1.3
pH = 8.89
Answer:
<em>Well, Your best answer will be is 2H+ + 2OH- -> 2H2O but you have to reduce it to H+ + OH- -> H2O. </em><em>Good Luck!</em>
Answer:
The answer to your question is: letter D.
Explanation:
Double replacement reaction is when the cation of one reactant replaces the metal of the other reactant and vice versa.
a. 2SO2 + O2 —> 2S03 This is a combination reaction
b. Zn + Cu(NO3)2 → Zn(NO3)2 + Cu This is a single replacement reaction.
c. 2H2O2–> 2H2O + O2 This is a decomposition reaction
d . AgNO3 + NaCl → AgCl + NaNO3 This is a double replacement reaction
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
