Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
Answer:
b Different amounts of food samples were used.
Explanation:
The mass of the two samples needs to be the same in order for the test to be accurate.
Answer:
The first option: Strontium Fluorate.
Explanation:
because Fluorine and oxygen combines to make fluorate, Strontium stays the same.
p.s: i need help in geo and there's an exam tomorrow.
