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ANEK [815]
3 years 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:
Gekata [30.6K]3 years ago
3 0

Answer:

The percent of ethanol is 0.1093%

Explanation:

Given:

t = time = 10 s

I = current = 320 mA

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

n = number of electrons = 4

Molecular weight of ethanol = 46 g/mol

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

First, it is necessary to calculate the mass of ethanol:

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 equation of ideal gas:

V=\frac{nRT}{P}

Here:

T = 26°C = 299 K

P = 1 atm

Substituting values:

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

The percent of ethanol:

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

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MAVERICK [17]

The pH of water is given to be 10.74.

This means = pOH = 14 - pH = 3.26

pOH = -log[OH-]= 3.26

[OH-] = 0.00055 M / N

The concentration of H2SO4 is given to be = 0.02 N

volume of water = 1 L

So moles of OH- = number of equivalents of OH-

                     = concentration X volume (L) = 0.00055  X 1 = 0.00055

So number of equivalents of H2SO4 requried = 0.00055

Hence volume of 0.02 N H2SO4 required = Moles / Molarity

                     = 0.00055/0.02 = 0.0275 L = 27.5 mL


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3 years ago
In an acid/base titration where NaOH(aq) is the titrant and HCl(aq) is the analyte, what is true about the moles of each reactan
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\huge{\color{magenta}{\fcolorbox{magenta}{black}{\huge{\color{white}{\fcolorbox{aqua}{black}{✿ᴀɴsᴡᴇʀ✿}}}}}}

<u>A</u><u>)</u><u> </u><u>The moles of NaOH are equal to moles of HCl.</u>

Explanation:

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Draw the lewis formula for chlorobenzene, a benzene derivative. include all hydrogen atoms and lone pairs.
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To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of H₂ and of O₂ and wants to use up both tanks to form 2
Lapatulllka [165]

the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

Given , to study a key fuel-cell reaction , a chemical engineer has 20.0L tanks of H2 and O2 and wants to use up both tanks to form 28.0mol of water at 23.8°C .

the reaction of the fuel cell is given by ,

H2(g) +1/2 O2 ( g) →H2O

Moles of H2 required = 28 mol

moles of O2 required = 14mol

Now according to van der waals equation ,

p= (nRT/ V-nb)  - an^2 / V^2

for H2 , a = 0.2453L^2bar /mol^2 , b= 0.02651L/mol

P for H2

= (28×0.0821×296.8/20-28×0.02651 ) -(0.2456 ×28×28/20×20)

P for H2 = 35.4291079-0.481376 = 34.9477319 atm

for O2 ,  a = 1.38L^2bar/mol^2 , b = 0.03186L/mol

P for O2 = (14×0.0821×296.8 /20-14×0.03186) - (1.382×14×14/20×20)

P for O2 = 17.4461807 - 0.67718 =16.7690007 atm

Hence , the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

Learn more about pressure here :

brainly.com/question/25965960

#SPJ4

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