2.37 x 0.004 Sig Fig
2 answers:
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Explanation:
Total volume of the solution is as follows.
19 ml + 19 ml = 38 ml
As, density is the mass divided by volume.
Mathematically, Density =
Now, calculate the mass of solution as follows.
mass of solution = Density × volume
= 1.00 g/ml × 38 ml
= 38 g
Specific heat of the solution is 4.184 J/g°C.
Relation between heat energy, mass, specific heat and temperature change temperature as follows.
Q =
=
= 1701.21 J
Now, milimoles = molarity × volume
=
= 26.6 mmol
Enthalpy of neutralization is as follows.
= 63.95 J/mmol
or, = 64 J/mmol
Thus, we can conclude that the enthalpy of neutralization in Joules/mmol for given solution is 64 J/mmol.
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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