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rjkz [21]
3 years ago
11

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 64. g of hexane is mi

xed with 72.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Nadusha1986 [10]3 years ago
8 0

Answer:

45.7g of CO₂ is the maximum amount of carbon dioxide that could be produced.

Explanation:

Hexane, C₆H₁₄, reacts with O₂ to produce H₂O and CO₂ as follows:

C₆H₁₄ + 13/2O₂ → 7H₂O + 3CO₂

<em>Where 1 mole of hexane reacts with 13/2 moles of O₂.</em>

<em> </em>

To solve this question we need to convert the mass of each reactant to moles in order to find the limiting reactant. With limiting reactant we can find the maximum amount of CO₂ produced:

<em>Moles C₆H₁₄ -Molar mass: 86.18g/mol-:</em>

64g C₆H₁₄ * (1mol / 86.18g) = 0.7426 moles C₆H₁₄

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

72.0g O₂ * (1mol / 32g) = 2.25 moles O₂

For a complete reaction of 2.25 moles of O₂ are required:

2.25 moles O₂ * (1mol C₆H₁₄ / 13/2 moles O₂) = 0.346 moles of C₆H₁₄

As there are 0.7426 moles of C₆H₁₄, C₆H₁₄ <em>is the excess reactant </em>and:

<em>Oxygen is limiting reactant.</em>

<em />

The maximum moles of CO₂ that could be produced are:

2.25 moles O₂ * (3mol CO₂ / 13/2 moles O₂) = 1.038 moles CO₂

<em>In grams:</em>

1.038 moles CO₂ * (44.01g / mol) =

<h3>45.7g of CO₂ is the maximum amount of carbon dioxide that could be produced.</h3>
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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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In the past, investigators would look at which two types of cues when trying to determine if someone was lying?
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Verbal and nonverbal.

They would look for shakiness of the voice (verbal) or darting eyes (nonverbal), and other things to that same effect.
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Answer:

Explanation:

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The value of the equilibrium constant for a reaction
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Answer:

a. changes with temperature.

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the equilibrium constant in terms of the Gibbs free energy of reaction and the temperature of the system:

lnK=-\frac{\Delta G}{RT}

It is possible to figure out that the equilibrium constant varies as temperature does, not only on the aforementioned definition, but also in the Gibbs free energy as it is also temperature-dependent. Therefore, the appropriate answer is a. changes with temperature.

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Answer:

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Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.

Explanation:

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