Raising of the boiling point is a colligative property. That means that it depends on the number of particles dissolved. The greater the number of particles the greater the increase in the boiling point. So, you can compare the effect of these solutes in the increase of the boiling point by writing the chemical equations and comparing the number of particles dissolved: 1)ionic lithium chloride, LiCl(s) --> Li(+) + Cl (-) => 2 ions; 2) ionic sodium chloride, NaCl(s) --> Na(+) + Cl(-) => 2 ions; 3) molecular sucrose, C12H22O11 (s) ---> C12H22O11(aq) => 1 molecule; 4) ionic phosphate, Na3PO4 --> 3Na(+) + PO4 (3-) => 4 ions; 5) ionic magnesium bromide, MgBr2 --> Mg(2+) + 2 Br(-) => 3 ions. <span>So, ionic phosphate produces the greatest number of particles and it will cause the greatest increase of the boiling point.</span><span />
The percentage yield obtained from the given reaction above is 74.8%
<h3>Balanced equation </h3>
P₄ + 6Cl₂ → 4PCl₃
Molar mass of P₄ = 31 × 4 = 124 g/mol
Mass of P₄ from the balanced equation = 1 × 124 = 124 g
Molar mass of PCl₃ = 31 + (35.5×3) = 137.5 g/mol
Mass of PCl₃ from the balanced equation = 4 × 137.5 = 550 g
<h3>SUMMARY</h3>
From the balanced equation above,
124 g of P₄ reacted to produce 550 g of PCl₃
<h3>How to determine the theoretical yield </h3>
From the balanced equation above,
124 g of P₄ reacted to produce 550 g of PCl₃
Therefore,
79.12 g of P₄ will react to produce = (79.12 × 550) / 124 = 350.9 g of PCl₃
<h3>How to determine the percentage yield </h3>
- Actual yield of PCl₃ = 262.6 g
- Theoretical yield of PCl₃ = 350.9 g
Percentage yield = (Actual /Theoretical) × 100
Percentage yield = (262.6 / 350.9) × 100
Percentage yield = 74.8%
Learn more about stoichiometry:
brainly.com/question/14735801
Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.
To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,
Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =
<em> 0.56 g Ca</em>
Answer:
16.8 g of AgCl are produced
Explanation:
The reactants are: NaCl and AgNO₃
The products are: AgCl, NaNO₃
Balanced equation: NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s) ↓
We convert the mass of AgNO₃ to moles → 10 g / 85g/mol = 0.117 moles
Ratio is 1:1, therefore 0.117 moles of nitrate will produce 0.117 moles of AgCl.
According to stoichiormetry.
We convert the moles to mass → 0.117 mol . 143.3g /1mol = 16.8 g
