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irakobra
1 month ago
5

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

ons in the correct order. Make sure the units of adjacent conversion factors cancel out. 1 L Octane times times times times times = 2171 g Carbon Dioxide
Chemistry
1 answer:
Tems11 [2.7K]1 month ago
7 0

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

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Thus, the rotation of the racemic mixture will be equal to 0°.


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1 month ago
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(c) Cu + S → CuS is classified as a redox reaction

Explanation:

The following reactions are presented:

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redox reactions

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1 month ago
An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver c
Tems11 [2777]
Q is determined to be 12.38. The Nernst equation is expressed as Ecell = E°cell - (2.303RT/nF) log Q, where Q represents the reaction quotient. The reaction quotient Q is calculated by taking the product of the products' concentrations divided by the product of the reactants' concentrations. For an electrochemical cell, Q is the concentration ratio of the solution at the anode compared to that at the cathode. Consequently, Q = [anode]/[cathode], specifically Q = 0.052/0.0042, arriving at a value of Q = 12.38.
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castortr0y [3046]
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1 month ago
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VMariaS [2998]

Answer:

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The controlled variable remains constant throughout the experiment.

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