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ikadub [295]
3 years ago
10

Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to

three significant figures and include the appropriate units.
Chemistry
1 answer:
zhuklara [117]3 years ago
8 0

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

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1. How many moles of CO2 can be produced from a reaction of 10.0 moles CzHg?
mr_godi [17]

Answer:

20 moles of CO2 can be produced from a reaction of 10.0 moles C2H6

Explanation:

In this reaction -

2 moles of C₂H6 produces four molecules of Carbon dioxide (CO2)

So 1 mole of C₂H6 will produce \frac{4}{2} = 2 moles of Carbon dioxide (CO2)

Thus, 10 moles of C₂H6 will produce 2 * 10 = 20 moles of Carbon dioxide (CO2)

3 0
3 years ago
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Answer:

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Explanation:

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3 0
3 years ago
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5 0
3 years ago
Read 2 more answers
How many moles of O2 molecules should be supplied to burn 1 mol of CH4 molecules in a domestic furnace?
Sati [7]
The combustion of 1 mole of methane (CH4) in a domestic furnace requires 2 moles of O2 molecules, assuming the combustion was complete or ideal. To solve this problem, use stoichiometry of the reaction's balanced chemical equation:

CH4 + 2O2 --> CO2 + 2H2O

The ratio of CH4 to O2 in terms of moles is 1:2. So 1 mole of CH4 needs 2 moles of O2.
4 0
3 years ago
45. The following data was collected for 3 compounds:
Vika [28.1K]

Answer:

The three compounds are different compounds

Explanation:

The mass of Nitrogen that combines with 1 gram of Oxygen in  Compound A = 1.750 g

The mass of Nitrogen that combines with 1 gram of Oxygen in  Compound B  = 0.8750 g

The mass of Nitrogen that combines with 1 gram of Oxygen in  Compound C  = 0.4375 g

According to the law of multiple proportions, when atoms of two different elements react to form compounds, the masses of one of the elements that combines with a fixed mass of the other element are in small whole number ratios.

The ratio of the masses are;

Mass of Nitrogen in Compound B/(Mass of Nitrogen in Compound C =  0.8750/0.4375 = 2

Mass of Nitrogen in Compound A/(Mass of Nitrogen in Compound C =  1.750/0.4375= 4

Mass of Nitrogen in Compound A/(Mass of Nitrogen in Compound B =  1.750/0.8750= 2

Given that the masses of Nitrogen in the three compounds are in small whole number ratios, the three compounds, Compound A, Compound B, and Compound C are different compounds.

5 0
3 years ago
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