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solniwko [45]
3 years ago
13

Wine goes bad soon after opening because the ethanol (CH,CH,OH) in it reacts with oxygen gas (0.) from the air to form water (11

0) and acetic acid (CH,COOH), the main ingredient of vinegar. What mass of water is produced by the reaction of 4.59 g of ethanol Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
daser333 [38]3 years ago
5 0

<u>Answer:</u> The mass of water produced is 1.8 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of ethanol = 4.59 g

Molar mass of ethanol = 46.07 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{4.59g}{46.07g/mol}=0.0996mol

The chemical equation for the reaction of ethanol with oxygen gas follows:

CH_3CH_2OH+O_2\rightarrow CH_3COOH+H_2O

By Stoichiometry of the reaction;

1 mole of ethanol produces 1 mole of water.

So, 0.0996 moles of ethanol will produce = \frac{1}{1}\times 0.0996=0.0996mol of water.

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.0996 moles

Putting values in equation 1, we get:

0.0996mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0996mol\times 18g/mol)=1.8g

Hence, the mass of water produced is 1.8 grams

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Allushta [10]

This question is not complete, the complete question is;

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally at pH  = 5.0 ( pKa = 5.0)

Neglect the volume change.

Options:

a) 6.10

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Answer:

the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

Explanation:

Given the data in the question,

Initially pKa = pH; so ratio is 1:1

thus, 0.4 M acid and base

Now, moles of NaOH = molarity × volume = 0.5 × 10 = 5 mmol = 5 × 10⁻³ mol.

Going into 500 mL ( 0.5 L ) of solution

new molarity will be;

⇒ moles / volume = 5 × 10⁻³ / 0.5 = 0.01 M

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original concentration of acid = 0.4 - 0.01 = 0.39 M

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so

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= 5 + log ( 0.41/0.39)

= 5 + log ( 1.0512)

= 5 + 0.021

pH = 5.02

Therefore the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

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Answer:

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(c) Y electrode

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Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻

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<em>Classify the descriptions by whether they apply to the X or Y electrode. </em>

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<em>(c) electrons in the wire flow toward.</em> Electrons in the wire flow toward the cathode (Y electrode).

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soldier1979 [14.2K]

Answer:

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