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RUDIKE [14]
4 years ago
12

The approximate recommended daily dietary intake of the amino acid valine, c5h11no2, is 2.60 g for a 100 kg adult. what is this

requirement in moles?
Chemistry
2 answers:
Murrr4er [49]4 years ago
5 0
Answer is: <span>requirement of valine is 0,0222 mol for a100 kilograms adult.
</span>m(C₅H₁₁NO₂) = 2,60 g.
M(C₅H₁₁NO₂) = 5 · 12 + 11 · 1 + 1 · 14 + 2 · 16 · g/mol.
M(C₅H₁₁NO₂) = 117 g/mol.
n(C₅H₁₁NO₂) = m(C₅H₁₁NO₂) ÷ M(C₅H₁₁NO₂).
n(C₅H₁₁NO₂) = 2,60 g ÷ 117 g/mol.
n(C₅H₁₁NO₂) = 0,0222 mol.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
Vsevolod [243]4 years ago
5 0
To calculate the requirement in mole, we need to calculate the molar mass.  
C5H11NO2 = (12 * 5) + (1 * 11) + 14 + (16 *2) = 117
 Hence number of moles = mass/molar mass
 Mass = 2.6g and molar mass = 117
 Number of moles = 2.6/117 = 0.022
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Convert 7.60 x1021 molecules of CO to Liters.
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you are given the number of moles, from the given formula you can get the molar mass which is 12.00g/mol + 16 g/mol =28 g/mol. using the formula n=m/Mr rearrange the formula and make m subject, you thn have m=nMr therefore 7.60x 1021 mol x 28g/mol= m

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217268.8g x 1l/1000g = 217,269 litres

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A titration reaction requires 38.20 mL phosphoric acid solution to react with 71.00 mL of 0.348 mol/L calcium hydroxide to reach
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1a. The balanced equation for the reaction is:

<h3>3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O </h3>

1b. The number of mole of Ca(OH)₂ is 0.0247 mole  

1c. The number of mole of H₃PO₄ is 0.0165 mole.

1d. The concentration of H₃PO₄ is 0.432 mol/L

2. The new concentration of the H₃PO₄ solution is 0.0432 mol/L

<h3>1a. The balanced equation for the reaction</h3>

<u>3</u>Ca(OH)₂ + <u>2</u>H₃PO₄ —> Ca₃(PO₄)₂ + <u>6</u>H₂O

<h3>1b. Determination of the mole of Ca(OH)₂</h3>

Volume of Ca(OH)₂ = 71 mL = 71 / 1000 = 0.071 L

Concentration of Ca(OH)₂ = 0.348 mol/L

<h3>Mole of Ca(OH)₂ =? </h3>

Mole = Concentration × Volume

Mole = 0.348 × 0.071

<h3>Mole of Ca(OH)₂ = 0.0247 mole </h3>

<h3>1c. Determination of the mole of H₃PO₄. </h3>

3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄.

Therefore,

0.0247 moles of Ca(OH)₂ will react with = \frac{0.0247 * 2}{3} = 0.0165 mole of H₃PO₄.

Thus, the number of mole of H₃PO₄ is 0.0165 mole

<h3>1d. Determination of the concentration of H₃PO₄</h3>

Volume of H₃PO₄ = 38.20 mL = 38.20/ 1000 = 0.0382 L

Mole of H₃PO₄ = 0.0165 mole

<h3>Concentration of H₃PO₄ =?</h3>

Concentration = \frac{mole}{volume} \\\\Concentration = \frac{0.0165}{0.0382}

<h3>Concentration of H₃PO₄ = 0.432 mol/L</h3>

<h3>2. Determination of the new concentration of the H₃PO₄ solution.</h3>

Initial Volume (V₁) = 10 mL

Initial concentration (C₁) = 0.432 mol/L

New volume (V₂) = 100 mL

<h3>New concentration (C₂) =?</h3>

The new concentration of the H₃PO₄ solution can be obtained as follow:

<h3>C₁V₁ = C₂V₂</h3>

0.432 × 10 = C₂ × 100

4.32 = C₂ × 100

Divide both side by 100

C₂ = \frac{4.32}{100}\\

<h3>C₂ = 0.0432 mol/L</h3>

Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L

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