Answer:
Explanation:
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In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:
In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:
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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>
Learn more about saturated solutions here:
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b. =
= 0,05 (mol)
⇒=
=0,05 mol
⇒= 0,05×135=6,75 (g)
c. =2×
=0,1 (mol)
⇒= 0,1×36,5= 3,65 (g)
⇒=
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=×100=
×100≈ 16,67%