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:

The moles of ethanol:

Applying the ideal gas law formula:

Here:
T = 26°C = 299 K
P = 1 atm
Substituting in the values:

The percentage of ethanol:
%
Thanks for the responses ッ. (By the way, the answers are located at the bottom of the question if anyone hasn’t noticed them.
Answer:
Explanation:
0.5678 G X GRAMS
KHC8H4O4 + NaOH = NaKC8H4O4 + H2O
1 MOL 1 MOL
0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4
0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE KHC8H4O4=0.00278 MOL NaOH
0.00278 MOL NaOH/26.26ml=0.106 molar