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Leni [432]
3 years ago
14

When 4-ketopentenoic acid is analyzed by combustion, it is found that a 0.3000g sample produces 0.579 g of CO2 and 0.142 g of H2

O. The acid contains only carbon, hydrogen, and oxygen. What is the empirical formula of the acid
Chemistry
1 answer:
SOVA2 [1]3 years ago
3 0

Answer:

C5H6O3

Explanation:

Number of moles of hydrogen = 0.142 * 2 /18  = 0.0158 moles of H

Number of moles of carbon = 0.579 * 1/44 = 0.013 moles of C

Mass of C = 0.013 moles of C * 12 = 0.156 g

Mass of H = 0.0158 moles of H * 1 = 0.0158 g

Mass of oxygen = 0.3000 - (0.156  + 0.0158 )

Mass of oxygen = 0.1282 g

Number of moles of oxygen =  0.1282 g/16 g/mol = 0.008 moles of O

Dividing through by the lowest number of moles

C-0.013/0.008 , H- 0.0158 /0.008 ,  O- 0.008/0.008

1.625,  1.975, 1 (multiply through by 3)

5,          6,      3

The required empirical formula is C5H6O3

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
enot [183]

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 $\large \boxed{\textbf{21.6 L}}$}

7 0
2 years ago
Chemistry help please! I just need to make sure the answers correct.... Thank you!
Veronika [31]
1. At constant tempaerature and pressure, 3 tablets produce 600cm^3 of gas
Thus calculating for 1 tablet that produces 600 / 3 = 200 cm^3
So now two tablets produce 200 x 2 = 400 cm^3
2. We have the equation PV = nRT, n being the number of moles
Pressure P = 1,000 kPa
Volume V = 3 L
R = 8.31 L kPa/mol-K
Temperature T = 298 K
n = PV / RT = (1000 x 3) / (8.31 x 298) = 3000 / 2476.38 = 1.21 moles
Number of moles = 1.21 moles.
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Explanation:

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