PH scale is used to determine how acidic, basic or neutral a solution is
pH can be calculated using the H₃O⁺
ph can be calculated as follows
pH = - log[ H₃O⁺]
[H₃O⁺] = 1 x 10⁻⁹
pH = - log [1 x 10⁻⁹]
pH = 9
pH of solution is 9
Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;
where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;
Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
The correct answer to
the question that is stated above is letter c, <span> the outer electron shell.</span>
Valence electrons occur<span> in the outermost shells of an </span>atom.
>> <span>Valence electrons are </span>electrons<span> that are associated with an </span>atom<span>, and that can participate in the formation of a </span>chemical bond.
Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
