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Setler79 [48]
1 year ago
9

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

Chemistry
1 answer:
rodikova [14]1 year ago
8 0

The volume of carbon dioxide produced is : 5.4 L

<u>Given data:</u>

moles of propane = 0.080 moles

<h3>Determine the volume of Carbon dioxide produced </h3>

The chemical reaction

C₃H₈  + 5O₂  ---- > 3CO₂ + 4H₂O

From the reaction

I mole of C₃H₈ = 3CO₂

0.080 moles of C₃H₈ =  3 * 0.080 = 0.24

Applying the equation below to determine the volume of CO₂

Pv = nRT

    = 0.24 * R * T

v = ( 0.24 * 8.314 * 273 ) / 1 atm

  = 544.7 ml = 5.4 L

Hence we can conclude that The volume of carbon dioxide produced is : 5.4 L

Learn more about Propane burning at STP : brainly.com/question/11903456

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What is the luster of aluminum
Mamont248 [21]

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Luster of aluminum is Silver-White.

Explanation:

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2 years ago
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
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<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

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             e^{-2.3346}=0.124 \gamma

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             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

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