Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)
moles KOH = 4 x 2 / 1
= 8 moles of KOH
molar mass KOH = 56 g/mol
mass of KOH = n x mm
mass of KOH = 8 x 56
= 448 g of KOH
hope this helps!
Answer:
A producer is a plant or fungi but it mainly is a organism that produces food for a primary consumer
Explanation:
Answer:56%
Explanation:
In the dewpoint chart when you line it up it ends up at 56%
the answer is d. when more solute is added
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6