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cestrela7 [59]
2 years ago
6

Acetaminophen (C8H9NO2) is the active ingredient in many nonprescription pain relievers. Each tablet contains 500 mg of acetamin

ophen, and a typical adult dose is two tablets every eight hours.
• Determine the molar mass of acetaminophen (show your work).
• Calculate the number of moles of acetaminophen in a single tablet (show your work).
• Calculate the number of moles of acetaminophen that an adult would have taken if she took three doses of acetaminophen in one day (show your work).
Chemistry
1 answer:
ohaa [14]2 years ago
7 0
1) Molar mass C8H9NO2

Element    Atomic mass    # of atoms   mass
                       g/mol                                  g
C               12                         8               12*8 = 96
H                 1                         9                1*9 =    9
N               14                         1               14*1 = 14
O               16                        2                16*2 = 32

                              molar mass =   96 + 9 + 14 + 32 = 151 g/mol

2) Number of mols in a tablet

# of moles = mass / molar mass = 0.500 g / 151 g/mol = 0.003311 moles

3) 3 doses * 2 tablets * 0.003311 moles / tablet = 0.020 moles
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Anyone? Please help
snow_tiger [21]

Answer:

The limiting reacting is O2

Explanation:

Step 1: data given

Number of moles O2 = 21 moles

Number of moles C6H6O = 4.0 moles

Step 2: The balanced equation

C6H6O + 7O2 → 6CO2 + 3H2O

Step 3: Calculate the limiting reactant

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed (21 moles).

C6H6O is in excess.

For 7 moles O2 we need 1 mol C6H6O

For 21 moles O2 we'll need 21/7 = 3 moles C6H6O

There will remain 4.0 - 3.0 = 1 mol C6H6O

Step 4: calculate products

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2

For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O

The limiting reacting is O2

7 0
3 years ago
Ñspent fuelî that can no longer be used to create energy.<br> nuclear waste ?
svetoff [14.1K]
Spent fuel that can no longer be used to create energy is waste. 
6 0
3 years ago
2. A chemical analysis of a sample provides the following elemental data:
Vadim26 [7]

Answer:

C3 H6 O2

Explanation:

first divide their mass by their respective molar mass, we get:

30.4 moles of C

61.2 moles of H

20.25 moles of O

now divide everyone by the smallest one of them then we get

C= 1.5

H= 3

O= 1

since our answer of C is not near to any whole number so we will multiply all of them by 2

so,

C3 H6 O2 is our answer

3 0
1 year ago
Boron has an average mass of 10.81. One isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. The
Andrej [43]

The average mass of an atom is calculated with the formula:

average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ...  an so on

For the boron we have two isotopes, so the formula will become:

average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)

We plug in the values:

10.81 = 0.1980 × 10.012938  + 0.8020 × mass of isotope (2)

10.81 = 1.98 + 0.8020 × mass of isotope (2)

10.81 - 1.98 = 0.8020 × mass of isotope (2)

8.83 = 0.8020 × mass of isotope (2)

mass of isotope (2) = 8.83 / 0.8020

mass of isotope (2) = 11.009975

mass of isotope (1) = 10.012938 (given by the question)

5 0
3 years ago
A graduated cylinder was filled with 25.0 mL of water. A solid object was immersed in the cylinder, raising the water level to 3
yawa3891 [41]

Answer:

Density of the object is 8759.494 grams/L

Explanation:

As we know density of an object is mass of the object divided by its volume

Given

Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.

Hence, volume of object is equal to

32.9 -25 = 7.9 mL

Mass of the object is 69.2 grams

Density

= \frac{69.2}{7.9 *10^{-3}}\\= 8759.494 grams/L

Density of the object is 8759.494 grams/L

7 0
3 years ago
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