We know that:
Molar Mass H2O: 18 g/mol
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>
<span>Using formula:
V= [mole fraction x molar mass] / density </span>
<span>mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 </span>
<span>morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 </span>
<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>
Yotal volume = 30 mL; therefore,
<span>0.0442 = (volume eugenol/30) </span>
<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
= 1.44/1.05
= about 1.37 mL </span>
The formula for mole fraction is:
-(1)
The solubility of oxygen gas = 1.0 mmol/L (given)
1.0 mmol/L means 1.0 mmol are present in 1 L.
Converting mmol to mol:
So, moles of oxygen = 0.001 mol
For moles of water:
1 L of water = 1000 mL of water
Since, the density of water is 1.0 g/mL.
So, the mass of water is 1000 g.
Molar mass of water = 18 g/mol.
Number of moles of water =
Substituting the values in formula (1):
Hence, the mole fraction is .
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
Answer:
The percent by mass of sodium chloride in the broth is 0.1875 %
Explanation:
Our first step is transforming the sodium chloride figure from mg to g
450mg x 1g/1000 mg= 0.45g
Now, we must calculate the percentage by using the following formula:
0.45g / 240g x 100= 0.1875%
Henceforth, we can conclude that the percentage by mass of sodium chloride in the broth is 0.1875%