Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four
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Answer:
26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃
Explanation:
To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?
moles of O₂= 26.25 ≅ 26.3
<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>
Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the
amount amount of solute that will be deposited is 1,927.413 grams.
<h3>How can the amount of solute deposited be found?</h3>
The volume of water 1.33 dm³ of water 70 °C.
The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles
At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles
Therefore;
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;
1.33 dm³ contains 2.25 moles.
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;
1.33 dm³ contains 0.53 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles
The number of moles that precipitate out = The amount of solute deposited
Which gives;
Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles
The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g
The molar mass of Pb(NO₃)₂ = 331 g/mol
The amount of solute deposited = Number of moles × Molar mass
Which gives;
The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>
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