Answer:
pH 8.89
Explanation:
English Translation
If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.
Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹
Assuming all of the salts involved all ionize completely
MgCl₂ ionizes to give Mg²⁺ and Cl⁻
MgCl₂ ⇌ Mg²⁺ + 2Cl⁻
1 mole of MgCl₂ gives 1 moles of Mg²⁺
Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M
Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻
Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²
(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²
[OH⁻]² = (6×10⁻¹¹)
[OH⁻] = √(6×10⁻¹¹)
[OH⁻] = 0.000007746 M
p(OH) = - log [OH⁻] = - log (0.000007746)
pOH = 5.11
pH + pOH = 14
pH = 14 - pOH = 14 - 5.11 = 8.89
Hope this Helps!!!
Answer:
82500000000000000000000000
Explanation:
This is the only answer I can come up with.
Answer: Strictly a laboratory analysis and can only be done using the data obtained during analysis
Explanation:
To find a solution to this problem, you need to use the data collected during the lab work. A guide could be finding the possible forms of hydrated copper chlorides in reference books. Since it's also a lab work, you can definitely compare your data with lab mates.
The formula CuxCly.zH₂O and its name chloride hydrate already gives you an idea of the possibilities of the value of the integers, hence you can take a good guess for the identity of the unknown salt and calculate the theoretical formular weight for it. From the that you can proceed to also find the mass of water and copper from your lab analysis.
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B)Buenos Aires was the city near the epicenter :)