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Lesechka [4]
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
Chemistry
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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3 years ago
Read 2 more answers
Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolvin
Sergio [31]

Answer:

\boxed{2.65}

Explanation:

1. Mass of acetylsalicylic acid (ASA)

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

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}

3. Concentration of ASA

c = \dfrac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}

4. Set up an ICE table

\begin{array}{ccccccc}\text{HA} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, &\text{H$_{3}$O$^{+}$} & + &\text{A}^{-}\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\

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

6. Solve the quadratic equation.

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

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258

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}}

4 0
3 years ago
What is the volume at STP of 3.44 x 1023 molecules of CO2
almond37 [142]

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains 6.022 \times 10^{23} molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}} (1b)

Where:

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.

If we know that P = 100000\,Pa, r = 3.44\times 10^{23}, N_{A} = 6.022\times 10^{23}, T = 273.15\,K and R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}, then the volume of carbon dioxide at STP is:

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)}

V = 12.972\,L

Therefore, the correct answer is C.

8 0
3 years ago
Please helpWhat’s the pH of a solution of ammonia that has a concentration of 0.335 M? The Kb of ammonia is
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

Ka = 1 * 10^-14/Kb

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

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

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

x=0.000014 M

Now;

pH = -log 0.000014 M

pH = 4.85

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

#SPJ1

7 0
2 years ago
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