The reactants are aluminum and iron nitrate.
Answer:
5.00 grams of salt contain more particles than 5.0 grams of sugar
Explanation:
Salt = NaCl
Molar mass = 58.45 g/mol
Sugar = C₁₂H₂₂O₁₁
Molar mass = 342.3 g/mol
Sugar's molar mass is higher than salt.
So 1 mol of sugar weighs more than 1 mol of salt
But 5 grams of salt occupies more mole than 5 grams of sugar
5 grams of salt = 5g / 58.45 g/m = 0.085 moles
5 grams of sugar = 5g/ 342.3 g/m = 0.014 moles
In conclusion, we have more moles of salt in 5 grams; therefore there are more particles than in 5 g of sugar.
Answer:
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Explanation:
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
