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mafiozo [28]
3 years ago
11

his is the chemical formula for epinephrine (the main ingredient in adrenaline): C9H13O3N A biochemist has determined by measure

ments that there are 33. moles of carbon in a sample of epinephrine. How many moles of nitrogen are in the sample?
Chemistry
1 answer:
Len [333]3 years ago
5 0

Answer:

3.67 moles of N

Explanation:

The epinephrine's chemical formula is: C₉H₁₃O₃N

We were told that a chemist found that in a mesaure of epinephrine, he found 33 moles of C

We must know that 9 moles of C are in 1 mol of C₉H₁₃O₃N so, let's make a rule of three:

If 9 moles of C are found in 1 mol of C₉H₁₃O₃N

Therefore 33 moles of C must be found in (33 .1) / 9 = 3.67 moles of C₉H₁₃O₃N

There is a second rule of three, then.

In 1 mol of C₉H₁₃O₃N we have 1 mol of N

Then, 3.67 moles C₉H₁₃O₃N must have (3.67 . 1) / 1 = 3.67 moles of N

Remember 1 mol of C₉H₁₃O₃N has 9 moles of C, 13 moles of H, 3 moles of O and 1 mol of N

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The correct mathematical expression for finding the molar solubility (s) of calcium phosphate is
GalinKa [24]

Answer is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.

<span> Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] = 3s(Ca₃(PO₄)₂) = 3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.

s = ⁵√(Ksp ÷ 108).

6 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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Explanation:

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