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2 years ago

Determine the molality of an aqueous solution that is 16.5 percent urea by mass

Chemistry

1 answer:

yarga [219]2 years ago

7 0

Take a hypothetical sample of exactly 100 grams of the solution.

(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea

((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O

(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea

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What are the four factors that affect the natural selection?

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Explanation:

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The following conditions must be met before a standard calibration curve equation can be used to solve for the concentration of

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2 years ago

You need to use a dilute hydrochloric acid solution in an experiment. However, the only bottle of hydrochloric acid in your lab’

mr_godi [17]

The pH of the diluted HCl solution is 1.3.

Explanation:

Given:

The concentrated HCl solution of 8.0 M. The 1.5 mL of 8.0 M HCl is diluted with water to 250 mL volume.

To find:

The pH of the diluted HCl solution.

Solution

The concentration of the HCl solution before dilution = $M_1=8.0 M$
The volume of the HCl solution taken for dilution = $V_1=1.5 ml$
The concentration of the HCl solution after dilution = $M_2=?$
The volume of the HCl solution after dilution = $V_2=250 mL$

Using the Dilution equation:

$M_1V_1=M_2V_2\\8.0M\times 1.5ml=M_2\times 250 mL\\M_2=\frac{8.0M\times 1.5ml}{250 mL}=0.048 M$

The concentration of diluted HCl solution = 0.048 M

$HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)$

In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:

$[H^+]=1\times 0.048M=0.048 M$

The pH of the diluted HCl solution :

$pH=-\log [H^+]\\=-\log [0.048M]=1.18 \approx 1.3$

The pH of the diluted HCl solution is 1.3.

Learn more about the dilution equation here:

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2 years ago

Which of these is equal to 1 mole?

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