Answer:
0,051g of O₂ are consumed
Explanation:
The reaction of precipitation of Fe(II) is:
4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)
The moles of Fe(II) you have in 85ml of 0.075 M Fe(II) are:
0,085L×0,075M = 6,4x10⁻³ moles of Fe(II)
For the reaction, you require 4 moles of Fe(II) and 1 mol of O₂(g) to precipitate the iron. Thus, moles of O₂(g) you require are:
6,4x10⁻³ moles of Fe(II)× = 1,6x10⁻³ moles of O₂(g)
These moles are:
1,6x10⁻³ moles of O₂(g)× = <em>0,051g of O₂ are consumed</em>
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I hope it helps!
Answer:
You must equalize the pressure inside and outside the flask to determine the total because it keeps the water level the same.
(credits to "coursehero")
Explanation:
Na = 22.99 amu
Cl = 35.45 amu
NaCl = 22.99 + 35.45 => 58.44 g/mol
there needs to be more info
Fe2(SO4)3 + 6KOH -> 3K2SO4 + 2Fe(OH)3