Answer:
648.68 mg
Explanation:
The reaction that takes place is:
- FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
First we<u> calculate how many moles of each reactant were added</u>, using the <em>given volumes and concentrations</em>:
- FeCl₃ ⇒ 100 mL * 0.240 M = 24 mmol FeCl₃
- NaOH ⇒ 100 mL * 0.182 M = 18.2 mmol NaOH
24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.
Now we <u>calculate how many moles of Fe(OH)₃ are formed</u>, using the <em>moles of the limiting reactant</em>:
- 18.2 mmol NaOH *
= 6.07 mmol Fe(OH)₃
Finally we <u>convert 6.07 mmol Fe(OH)₃ to grams</u>, using its<em> molar mass</em>:
- 6.07 mmol Fe(OH)₃ * 106.867 mg/mmol = 648.68 mg
<span>Electrons, the negatively charged particles that carry electricity in metals, can be ejected from the material if it is bombarded by a large number of high-speed electrons. This effectively, leads to some metallic electrons being "knocked" out of the material. </span>
<span>A. Helium </span>atomic number 2
hope it helps
Answer:
39 mol AgNO3
Explanation:
We have the equation 4HNO3 + 3Ag -----> 3AgNO3 + NO + 2H2O
We want to calculate the number of silver nitrate (AgNO3) moles that would be produced from 52 moles of nitric acid ( HNO3 )
We can calculate this by using mole ratio as well as dimensional analysis.
The mole ratio of Silver nitrate to nitric acid based on the balanced equation is 3AgNO3:4HNO3.
Using this we can create a table: The table is attached.
Breakdown of the table.
The moles of nitric acid cancel out and we multiply 52 by 3/4 to get 39 moles of Silver nitrate.
