Answer:
2.15
Explanation:
For this question, we have to remember the <u>pH formula</u>:
![pH~=~-Log[H_3O^+]](https://tex.z-dn.net/?f=pH~%3D~-Log%5BH_3O%5E%2B%5D)
By definition, the pH value is calculated when we do the -Log of the concentration of the <u>hydronium ions</u> (
). So, the next step is the calculation of the <u>concentration</u> of the hydronium ions. For this, we have to use the <u>molarity formula</u>:

We already know the number of moles (0.0231 moles) and the volume (3.33 L). So, we can plug the values into the molarity formula:

With this value, now we can calculate the pH value:
![pH~=~-Log[0.00693~M]~=~2.15](https://tex.z-dn.net/?f=pH~%3D~-Log%5B0.00693~M%5D~%3D~2.15)
<u>The pH would be 2.15</u>
I hope it helps!
Answer:
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Explanation:
Mass of gold = m
Volume of gold = v
Surface area on which gold is plated = 
Thickness of the gold plating = h = 0.5 mm = 0.05 cm
1 mm = 0.1 cm

Density of the gold = 

Moles of gold = 

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :
of electrons
Number of electrons = N =
Charge on single electron = 
Total charge required = Q

Amount of current passes = I = 8 Ampere
Duration of time = T



It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide.
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂ - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present
so if 3 mol are present in 1000 ml
then volume for 0.038 mol = 1000/3 * 0.038
= 12.67 ml
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
Each compound will be neutral