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astraxan [27]
2 years ago
10

Write the molecular formula of glucose​

Chemistry
1 answer:
MariettaO [177]2 years ago
6 0
The molecular formula of glucose:
C₆H₁₂O₆
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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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2 years ago
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3 years ago
Complete the mechanism for the base-catalyzed opening of the epoxide below by adding any missing atoms, bonds, charges, nonbondi
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Complete Question:

check the first image for complete part of the question

Answer and Explanation:

Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.

Ring opening of epoxide with acid:  

In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:  

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)

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