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Gekata [30.6K]
3 years ago
8

Now it's time to make dinner. On the menu tonight are fried chicken, green beans and cornbread. (5) The first thing you need to

do is turn on the burners so that you can fry the chicken and boil the water for green beans. You have a natural-gas (methane, CH4) stove. When you ignite a natural gas stove, the methane reacts with oxygen (O2) to produce carbon dioxide and water vapor. The reaction is CH4 + O2 ----> CO2 + H2O. Explain what type of reaction this is.
Chemistry
1 answer:
ladessa [460]3 years ago
5 0

Answer:

Complete combustion

Explanation:

Let's consider the reaction that occurs when methane reacts with oxygen to produce carbon dioxide and water vapor.

CH₄ + O₂ → CO₂ + H₂O

The reaction between an organic compound with oxygen to form carbon dioxide and water vapor is known as complete combustion. If instead of carbon dioxide, the product were carbon monoxide or carbon, the combustion would be incomplete.

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A sample of O2 gas occupies a volume of 871 mL at 25 °C. If pressure remains constant, what would be the new volume if the tempe
Nookie1986 [14]

Answer:

\boxed{\text{(a) 783 L; (b) 900. mL; (c) 3.20 L}}

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The pressure and the number of moles are constant, so, to calculate the volume, we can use Charles' Law.

\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}

(a) Volume at -5 °C

Data:

V₁ = 871mL; T₁ =  25 °C

V₂ = ?;         T₂ =   -5 °C

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T₁ = ( 25 + 273.15) K = 298.15 K

T₂ =   (-5 + 273.15) K = 268.15 K

(ii) Calculate the new volume

\begin{array}{rcl}\dfrac{V_{1}}{T_{1}} &= &\dfrac{V_{2}}{T_{2}}\\\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{268.15}\\\\2.921 &= &\dfrac{V_{2}}{268.15}\\\\{ V_{2}} &=& 2.9210 \times 268.15\\&=& \textbf{783 mL}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{783 mL}}$}

(b) Volume at 95 °F

95 °F = (95 - 32) × 5/9 = 63 × 5/9 = 35 °C

35 °C = (35 + 273.15) K = 308.15 K

\begin{array}{rcl}\dfrac{871}{298.15} &= &\dfrac{V_{2}}{308.15}\\\\2.921 &= &\dfrac{V_{2}}{308.15}\\\\{ V_{2}} &=& 2.9210 \times 308.15\\&=& \textbf{900. mL}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{900. mL}}$}

(c) Volume at 1095 K

\begin{array}{rcl}\\\dfrac{871}{298.15} &= &\dfrac{V_{2}}{1095}\\\\2.921 &= &\dfrac{V_{2}}{1095}\\\\{ V_{2}} &=& 2.9210 \times 1095\\&=& \text{3200 mL}\\&=& \textbf{3.20 L}\\\end{array}\\\text{The gas will occupy $\large \boxed{\textbf{3.20 L}}$}

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