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4 years ago

15

A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

lutions. The first bottle is one-half full with a 20% alcohol solution, the second bottle is one-fifth full with a 30% alcohol solution, and the third bottle is one-tenth full with a 50% alcohol solution. The student pours the contents of all three bottles into an empty one-liter bottle. What is the approximate alcohol content of this mixed solution

1 answer:

Svetllana [295]4 years ago

3 0

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml. having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

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Answer:

$\boxed{2.65}$

Explanation:

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$m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}$

2. Moles of ASA

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3. Concentration of ASA

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4. Set up an ICE table

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5. Solve for x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0$

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$a = 1; b = 3.33 \times 10^{-4}; c = -5.851 \times 10^{-6}$

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7. Calculate the pH

$\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}$

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3 years ago

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almond37 [142]

Answer:

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$P$ - Pressure, measured in pascals.

$V$ - Volume, measured in liters.

$r$ - Amount of molecules, no unit.

$N_{A}$ - Avogadro's number, no unit.

$R_{u}$ - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

$T$ - Temperature, measured in Kelvin.

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$V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}$

$V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}$

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3 years ago

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natta225 [31]

From the calculation, the pH of the solution is 4.85.

<h3>What is the pH?</h3>

The pH is defined as the hydrogen ion concentration of the solution. We have the ICE table as;

HA + H2O ⇔ H3O^+ + A^-

I 0.335 0 0

C -x +x +x

E 0.335 - x x x

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Ka = 1 * 10^-14/1.8 × 10^–5

Ka = 5.56 * 10^-10

Ka = [H3O^+] [A^-]/[HA]

But [H3O^+] = [A^-] = x

5.56 * 10^-10 = x^2/ 0.335 - x

5.56 * 10^-10(0.335 - x ) = x^2

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x^2 + 5.56 * 10^-10x - 1.86 * 10^-10 = 0

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Learn more about pH:brainly.com/question/15289741

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7 0

2 years ago