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Aleksandr-060686 [28]
3 years ago
9

You are assigned the task of separating a desired granular material, with a density of 3.26 g/cm3, from an undesired granular ma

terial that has a density of 2.04 g/cm3. You want to do this by shaking the mixture in a liquid. A solid will float on any liquid that is more dense. Using an Internet-based source, find the densities of the following substances: carbon tetrachl oride, hexane, benzene, and diiodomethane. Which of these liquids will serve your purpose, assuming no chemical interaction between the liquid and the solids?
Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
5 0

Answer:

diiodomethane

Explanation:

The densities of each of the liquids mentioned in the question are stated below;

CCl4- 1.59 g/cm^3

Hexane- 0.672 g/cm^3

Benzene- 0.8765 g/cm^3

Diiodomethane- 3.3 g/cm^3

Clearly, the density of diiodomethane methane is almost the same as that of the desired granular material, hence the undesired granular material having a density of 2.04 g/cm3 will float in diiodomethane thus separating the two granular materials.

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Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of O_2 = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

PV=nRT

where,

P = pressure of gas

V = volume of gas

n = the number of moles

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R = gas constant = 0.0821 L atm/mole K   (Given)

By rearranging the above ideal gas law equation, we get

V=\frac{nRT}{P}

Now put all the given values in this expression, we get the value of volume.

V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L

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2 years ago
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3 0
3 years ago
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8
love history [14]

Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose  = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M

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