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spin [16.1K]
1 year ago
10

use dimensional analysis to determine how many atoms of o2 produced in a reaction that yields 2 mol of o2

Chemistry
1 answer:
Alex73 [517]1 year ago
8 0

The number of atoms of oxygen is 2.4 * 10^24.

<h3>How many atoms are produced?</h3>

We know that mole of a substance contains the Avogadro's number of atoms. The Avogadro's number gives the number of atoms, molecules, ions and atoms that could be found in one mole of a substance.

Now;

Given that ;

1 mole of oxygen contains two atoms of oxygen, the number of atoms that are contained in 2 moles of the oxygen molecule is;

2 * 2 * 6.023 * 10^23

= 2.4 * 10^24

Learn more about dimensional analysis:brainly.com/question/13078117

#SPJ1

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)
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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
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Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

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