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Anon25 [30]
3 years ago
15

When water is subjected to an electric current, it decomposes to hydrogen and oxygen gas. If 1.00 g of water is decomposed at 1.

00 atm and 25 °C, what volume of oxygen gas is collected?
Chemistry
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

Electrolysis is a process that __________.A. produces an electrical current while decomposing water into hydrogen gas and oxygen gas. B. requires an electrical current to decompose water into hydrogen gas and oxygen gas. C. produces heat energy while decomposing water into hydrogen gas and oxygen gas. D. requires heat energy to decompose water into hydrogen gas and oxygen gas.

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We are asked to provide an equation for the transformation of 2-phenylethanoic acid to 2-phenylethanol. This type of a reaction is converting a carboxylic acid to an alcohol, which is classified as a reduction reaction since we are decreasing the number of bonds to oxygen in the molecule. In order to reduce a carbonyl to an alcohol, we need a source of hydride, H⁻. Reducing the carboxylic acid once will convert it to the aldehyde. However, we need to reduce the functional group all the way down to an alcohol, which is another reduction step after aldehyde formation. Therefore, the hydride source of choice is lithium aluminum hydride, LiAlH₄.

A reaction scheme is provided to show the reaction of the reduction of carboxylic acid to alcohol. The first step is addition of lithium aluminum hydride which does the reduction, and the second step is a work-up of acid which protonates the alcohol to get the final product.

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of
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Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

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