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Alex73 [517]
3 years ago
13

The natural abundances of elements in the human body, expressed as percent by mass, are oxygen (o), 65 percent; carbon (c), 18 p

ercent; hydrogen (h), 10. Percent; nitrogen (n), 3.0 percent; calcium (ca), 1.6 percent; phosphorus (p), 1.2 percent; all other elements, 1.2 percent. Enter your answers in scientific notation. Calculate the mass in grams of each element in the body of a 64−kg person.
Chemistry
2 answers:
mr Goodwill [35]3 years ago
6 0

Answer:

O: 4.2 × 10⁴ g

C: 1.2 × 10⁴ g

H: 6.4 × 10³ g

N: 1.9 × 10³ g

Ca: 1.0 × 10³ g

P: 7.7 × 10² g

Other elements: 7.7 × 10² g

Explanation:

Natural abundances of elements in the human body (in percent by mass)

  • oxygen (O), 65%
  • carbon (C), 18%
  • hydrogen (H), 10%
  • nitrogen (N), 3.0%
  • calcium (Ca), 1.6%
  • phosphorus (P), 1.2%
  • all other elements, 1.2%

The mass of a body is 64 kg = 64 × 10³ g = 6.4 × 10⁴ g. We can calculate the mass of each element by <em>multiplying the body weight by the mass percent of each element</em>.

O: 6.4 × 10⁴ g × 65% = 4.2 × 10⁴ g

C: 6.4 × 10⁴ g × 18% = 1.2 × 10⁴ g

H: 6.4 × 10⁴ g × 10% = 6.4 × 10³ g

N: 6.4 × 10⁴ g × 3.0% = 1.9 × 10³ g

Ca: 6.4 × 10⁴ g × 1.6% = 1.0 × 10³ g

P: 6.4 × 10⁴ g × 1.2% = 7.7 × 10² g

Other elements: 6.4 × 10⁴ g × 1.2% = 7.7 × 10² g

Vlada [557]3 years ago
3 0

It is given that the person weighs 62 kg = 62,000 g

Natural abundances in mass percent are:

O = 65%

C = 18%

H = 10%

N = 3.0%

Ca = 1.6%

P = 1.2%

Corresponding weights of the elements are:

O = 65/100 * 62000 g = 40.30 * 10^3 g

C = 18/100 * 62000 g = 11.16 * 10^3 g

H = 10/100 * 62000 g = 62.00 * 10^2 g

N = 3.0/100 * 62000 g = 18.60 * 10^2 g

Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g

P = 1.2/100 * 62000 g = 7.44 * 10^2 g


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1 year ago
A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa
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<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

The chemical equation for the reaction of NaOH and monoprotic acid follows:

NaOH+HX\rightarrow NaX+H_2O

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = \frac{0.0225}{2}=0.01125moles

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

  • <u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

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Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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