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Mars2501 [29]
3 years ago
10

A solution is 5.00% by volume of ethanol dissolved in water. How many mL of ethanol are in 500 mL of the solution?

Chemistry
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

v_{solute}=25mL

Explanation:

Hello there!

In this case, according to the by-volume concentration of this solution, it is possible for us to use its mathematical definition as shown below:

\%v=\frac{v_{solute}}{v_{solution}}*100\%

Thus, given the percent and the volume of the solution, we can solve for the volume of ethanol (solute) as shown below:

v_{solute}=\frac{\%v*v_{solution}}{100\% }\\\\v_{solute}=\frac{5.00\%*500mL}{100\%}\\\\ v_{solute}=25mL

Best regards!

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Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:CH₃CH₂OH + Cr
garik1379 [7]

Balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The chemical equation

CH₃CH₂OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺

First assign the oxidation number for each atom in the equation.

\overset{+3}{C}\overset{+1}{H_3} \overset{-1}{C} \overset{+1}{H_2} \overset{-2}{O} \overset{+1}{H} + \overset{+6}{Cr_2} \overset{-2}{O_7} + \overset{+1}{H} \rightarrow \overset{+3}{C}\overset{+1}{H_3} \overset{+3}{C}\overset{-2}{O} \overset{-2}{O} \overset{+1}{H} + \overset{+3}{Cr}

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻

Reduction: Cr₂O₇ + 6e⁻  → 2Cr⁺³

Now, balance the charge

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³

Now balance the oxygen atoms

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O

Now, make electron gain equivalent to lost

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺ }  × 3

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O } × 2

Now,

Oxidation: 3C₂H₆O + 3H₂O → 3C₂H₄O₂ + 12e⁻ + 12H⁺

Reduction: 2Cr₂O₇ + 12e⁻ + 28H⁺ → 4Cr⁺³ + 14H₂O

Now, add the both equations

3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Thus from the above conclusion we can say that the balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

#SPJ4

7 0
1 year ago
PLEASE HELPPP ._. xd
Juli2301 [7.4K]
1. Hyaline: most common; found in the ribs, nose, trachea. Is a precursor of bone
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6 0
3 years ago
Two colorless solutions are mixed together bubbles form as the solution is stirred give to possible explanations for this result
Aleks [24]
One possible answer could be that a chemical reaction has occurred.
5 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
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