How to get a hot bf quick?

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

vichka [17]

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

3 0

4 years ago

The specific heat of water is 4.184 J/ g°C, the specific heat of wood is 1.760 J/g°C, and the specific heat of coal is 1.26 J/g°

Tomtit [17]

Answer:

D. Water

Explanation:

Specific heat is defined as the amount of heat you need to add to a substance per unit of mass to increase its temperature in 1°C.

That means if you have the same mass of substances, the one with the higherst specific heat is the one that requires the most amount of heat to increase its temperature by one degree Celsius.

As you can see, water is the material with the highest specific heat doing the material that requires the most heat.

Right answer:

<h3>D. Water </h3>

6 0

3 years ago

When is di- used in the name of a hydrocarbon?

harkovskaia [24]

Di- is used when you are naming organic compounds. If you have the same substituent repeated twice in the compund
For example: CH3-CH(CH3)-CH2-CH(CH3)-CH3
This will be named 2,4-dimethylpentane

6 0

3 years ago

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon

Varvara68 [4.7K]

Answer: 31.8 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles$

$\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles$

$Al_2O_3+3C\rightarrow 2Al+3CO$

According to stoichiometry :

1 mole of $Al_2O_3$ require 3 moles of $C$

Thus 0.59 moles of $Al_2O_3$ will require= $\frac{3}{1}\times 0.59=1.77moles$ of $C$

Thus $Al_2O_3$ is the limiting reagent as it limits the formation of product and $C$ is the excess reagent as it is present in more amount than required.

As 1 mole of $Al_2O_3$ give = 2 moles of $Al$