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:
The moles of ammonia present in a 1.284 g sample are 0.075.
Explanation:
Molar mass is the mass of one mole of a substance, which can be an element or a compound. If the molar mass of ammonia is 17,030 g / mol, then in 1 mole of ammonia there are 17,030 g.
So, in this case, the following rule of three can be applied: if by definition of molar mass in there are 17.030 g in 1 mole of ammonia, 1.284 g of ammonia in how many moles will it be?
![moles=\frac{1.284 g*1 mole}{17.030 g}](https://tex.z-dn.net/?f=moles%3D%5Cfrac%7B1.284%20g%2A1%20mole%7D%7B17.030%20g%7D)
moles=0.075
<u><em>The moles of ammonia present in a 1.284 g sample are 0.075.</em></u>
Answer:
10 Kilograms
Explanation:
"millimetres" is a unit of measurement and is used for measuring distances.
"newton" is a unit of weight and is used to represent the weight of an object
:kilograms" is the one used for measuring mass
Answer:
zncl2. . . . . . . . . . . . . . . .