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bekas
4 months ago
11

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a

nswers as chemical formulas separated by a comma. nothing Request Answer Part B Cadaverine, a foul-smelling substance produced by the action of bacteria on meat, contains 58.55% C, 13.81% H, and 27.40% N by mass; its molar mass is 102.2 g/mol. Express your answers as chemical formulas separated by a comma. nothing Request Answer Part C Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains 59.0% C, 7.1% H, 26.2% O, and 7.7% N by mass; its molar mass is about 180 amu. Express your answers as chemical formulas separated by a comma. nothing
Chemistry
1 answer:
castortr0y [3K]4 months ago
7 0

Answer:

A) The molecular formula for ibuprofen isC_{13}H_{18}O_2

B) The molecular formula for Cadaverine is C_{5}H_{14}N_2

C) The molecular formula for Epinephrine is C_9H_{13}O_3N_1

Explanation:

Element percentage in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is =C_xH_yO_z

Count of carbon atoms in one ibuprofen molecule;

75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100

x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13

Count of hydrogen atoms in one ibuprofen molecule;

8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100

y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18

Count of oxygen atoms in one ibuprofen molecule;

15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100

z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2

Molecular formula for ibuprofen:

= C_xH_yO_z= C_{13}H_{18}O_2

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is =C_xH_yN_z

Count of carbon atoms in one Cadaverine molecule;

58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100

x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5

Count of hydrogen atoms in one Cadaverine molecule;

13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100

y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14

Count of nitrogen atoms in one Cadaverine molecule;

27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100

z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2

Molecular formula for Cadaverine:

= C_xH_yN_z= C_{5}H_{14}N_2

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is =C_xH_yO_zN_w

Count of carbon atoms in one Epinephrine molecule;

59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100

x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9

Count of hydrogen atoms in one Epinephrine molecule;

7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100

y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13

Count of oxygen atoms in one Epinephrine molecule;

26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100

z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3

Count of nitrogen atoms in one Epinephrine molecule;

7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100

w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1

Molecular formula for Epinephrine:

= C_xH_yO_zN_w= C_9H_{13}O_3N_1

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In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an ar
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Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in  15.3 \frac{atoms}{min-g}

The dissolution rate of the sample is given by 9.16 \frac{atoms} {min-gram}

The C-14 proportion within the sample can be determined as per = \frac {9.16}{15.3} \\\\ = 0.5987

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\

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Thus, the age of the sample is represented as = n \times\ half-life

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3 months ago
In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What
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Answer:

The force is 38503.5N.

Explanation:

From the problem, we determine:

P (pressure) = 5.00 atm.

Next, to find the force in Newtons (N), we must convert 5 atm into N/m², as shown:

1 atm equals 101325 N/m².

So, 5 atm equals 5 x 101325 = 506625 N/m².

A (the piston area) = 0.0760 m².

Pressure signifies force per unit area, mathematically represented as

P = F/A.

From this, we find F = P × A.

F = 506625 × 0.0760.

Therefore, F = 38503.5N.

Thus, the piston experiences a force of 38503.5N.

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