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3 years ago

Give two real life applications of chromatography

Chemistry

2 answers:

AleksAgata [21]3 years ago

7 0

Answer:

1. Additive detection

2. Crime scene testing

koban [17]3 years ago

4 0

Answer:

Food testing.

Beverage testing.

Drug testing.

Forensic testing.

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How many moles are present in 2.7 liters of Argon gas at STP?

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2 years ago

Which of the following has the strongest magnetic field?

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3 years ago

A 2.2 M solution is made by with 0.45 moles of a solute. What is the final volume of this solution?

Savatey [412]

Answer: The final volume of this solution is 0.204 L.

Explanation:

Given: Molarity of solution = 2.2 M

Moles of solute = 0.45 mol

Molarity is the number of moles of solute present divided by volume in liters.

$Molarity = \frac{no. of moles}{Volume (in L)}$

Substitute the values into above formula as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\2.2 M = \frac{0.45}{Volume}\\Volume = 0.204 L$

Thus, we can conclude that the final volume of this solution is 0.204 L.

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3 years ago

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3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago