Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
Answer:
E°= E°cathode- E° anode= 0.271-0.330= -0.59V
Explanation:
NB: the stoichiometry does not affect E°values,
And the more positive the E° values , the greater it's tendency to become spontaneous and hence irreversible, and the more negative the E° values the more likely to become less spontaneous and reversible, hence the above reaction is reversible
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don't know because this is the question which I never heard
Answer:
1.) 2H2 + O2 ---> 2H2O
2.) CH4 + 4Cl2 ---> CCl4 + 4HCl
Explanation:
In order to balance the equation, you have to make sure there are the same amount of each element on both sides.
Ex (#1). By adding the coefficients infront of H2 and H20, You now have 4 hydrogen's and 2 oxygens on the left side as well as 4 hydrogens and 2 oxygens on the right side.
Sorry if that's confusing, but hope this helps! :)
Answer:
experiment can show the movement of pollutants through the groundwater!
Explanation:
