Acetaminophen as a chemical formula of C8H9NO2. The molar
masses are:
C8H9NO2 = 151.163 g/mol
C = 12 g/mol
H = 1 g/mol
N = 14 g/mol
O = 16 g/mol
<span>TO get the mass percent, simply multiply the molar mass of
each elements with the number of the
element divide by the molar mass of acetaminophen, that is:</span>
%C = [(12 * 8) / 151.163] * 100% = 63.50%
%H = [(1 * 9) / 151.163] * 100% = 5.954%
%N = [(14 * 1) / 151.163] * 100% = 9.262%
<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>
Answer:

Explanation:
Hello,
On the attached document, you will find the procedure regarding the required volume. At first, it is necessary to know atoms by day produced, by using a rule of three. Subsequently, we compute the moles per day and finally the volume via the ideal gas equation at STP.
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Answer:
9.96*10^21
Explanation:
Molar mass of K2O=29*2+16
= 74g per mol
number of moles in the sample= 1.224/ 74
=0.1654
Number of particles in 1 mole=6.0221409*10^23
Number of particles= 0.01654*6.0221409*10^23
=9.96*10^21
To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.
For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm
For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L
For temperature,
T = 45 + 273 = 318 K
For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g
Assuming ideal gas,
PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol
Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>