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!!!
The solution to your problem is as follows:
<span>2 KClO3 → 2 KCl + 3 O2
</span><span>MW of KCL = 39.1 + 35.35 = 74.6 g/mole
62.6/74.6 = 0.839 moles KCl produced
0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
</span>
Therefore, there are <span>40.27g of O2 produced during the reaction.
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
The type of circuit depicted above is an example of a parallel circuit. Parallel circuits are circuits that have several pathways for current to flow; the easiest way to see this is by drawing a condensed diagram. If one of the bulbs were removed (that particular pathway was opened so electricity couldn’t flow), the current could simply go through another pathway using another lightbulb and the circuit would still be complete. Series circuits only have one pathway that the current can flow through, which this can’t be since it has multiple. The circuit isn’t shorted since the lightbulb reduces the voltage to zero, and the circuit is closed since the lightbulbs are on.
Hope this helps!
Answer:
8.3ml
Explanation:
to get volume u have to divide 25g over the density, i rounded to the nearest tenth, if you don't want to then write out the full number with all the decimals
I believe that #1 is the lie, but I'm not great at this subject.
