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Ede4ka
9 days ago
12

One type of breathalyzer employs a fuel cell to measure the quantity of alcohol in the breath. When a suspect blows into the bre

athalyzer, ethyl alcohol is oxidized to acetic acid at the anode:
CH3CH2OH(g)+4OH−(aq)→HC2H3O2(g)+3H2O(l)+4e−

At the cathode, oxygen is reduced:
O2(g)+2H2O(l)+4e−→4OH−(aq)

The overall reaction is the oxidation of ethyl alcohol to acetic acid and water. When a suspected drunk driver blows 186 mL of his breath through this breathalyzer, the breathalyzer produces an average of 320 mA of current for 10 s.

Required:
Assuming a pressure of 1.0 atm and a temperature of 26C, what percent (by volume) of the driver's breath is ethanol?
Chemistry
1 answer:
Alekssandra [2.6K]9 days ago
3 0

Response:The ethanol percentage is 0.1093%

Explanation:

As given:

t = time = 10 s

I = current = 320 mA

F = Faraday's constant = 96485.3365 C mol⁻¹

n = number of electrons transferred = 4

Molecular weight of ethanol is 46 g/mol

Question: What is the percent (by volume) of ethanol in the driver's breath, %E =?

First, calculate the ethanol mass:

W=\frac{\frac{46}{4} *0.32*10}{96485.3365} =3.814x10^{-4} g

The moles of ethanol:

n_{ethanol} =3.814x10^{-4} g*\frac{1mol}{46} =8.291x10^{-6} moles

Applying the ideal gas law formula:

V=\frac{nRT}{P}

Here:

T = 26°C = 299 K

P = 1 atm

Substituting in the values:

V=\frac{8.291x10^{-6}*0.082*299 }{1} =2.033x10^{-4} L=0.2033mL

The percentage of ethanol:

E=\frac{0.2033}{186} *100=0.1093%

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The ideal gas law tends to become inaccurate when Group of answer choices the pressure is raised and the temperature is lowered.
VMariaS [2631]

Answer: Option (a) is the correct answer.

Explanation:

Under conditions of low pressure and high temperature, gas molecules exhibit negligible attractions or repulsions among themselves. Hence, gases behave ideally in these scenarios.

Conversely, at low temperatures, there is a reduction in the kinetic energy of gas molecules, while high pressure compels the molecules to be closer together.

Thus, attractive forces emerge between molecules in conditions of low temperature and high pressure, causing gases to be termed real gases.

Therefore, we conclude that the ideal gas law becomes less accurate when pressure increases and temperature decreases.

5 0
15 days ago
A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
KiRa [2612]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g ×\frac{1mol}{159,609g} = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g ×\frac{1mol}{18,02g} = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0
1 month ago
A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
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Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product (CH3COOH)^{2} gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol (CH3COOH)^{2} / 2 mol CH3COOH)

= 0.005 mol (CH3COOH)^{2}

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

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