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Alex787 [66]
3 years ago
12

Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

in 195 g of chlorobenzene has a boiling point of 133.30 °C. Calculate Kbp for chlorobenzene based on this experiment.
Chemistry
1 answer:
vichka [17]3 years ago
3 0

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

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4 Al + 3 O2 --> 2 Al2O3
SIZIF [17.4K]

Answer:

The answer to your question is 7.4 moles of Aluminum

Explanation:

Data

moles of Al = ?

moles of Al₂O₃ = 3.7

Balanced chemical reaction

                4 Al  +  3 O₂  ⇒  2 Al₂O₃

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.

                4 moles of Aluminum ----------------- 2 moles of Al₂O₃

                 x                                  ----------------- 3.7 moles of Al₂O₃

                            x = (3.7 x 4) / 2

                            x = 14.8 / 2

                            x = 7.4 moles of Aluminum

5 0
3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

4 0
3 years ago
If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol
maks197457 [2]
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m -  molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
8 0
3 years ago
How was the work of Newlands similar to that of Mendeleev on the periodic table?
kiruha [24]
It was similar because  they both arranged the elements in order of increasing atomic mass.
8 0
3 years ago
Read 2 more answers
What shows the number of protons in it's nucleus
user100 [1]
The Atomic Number.

-Hope this helps!
8 0
3 years ago
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