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butalik [34]
3 years ago
6

Problem Page Combustion of hydrocarbons such as pentane ( C5 H12 ) produces carbon dioxide, a "greenhouse gas." Greenhouse gases

in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.(a) Write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water. (b) Suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 degree C. Calculate the volume of carbon dioxide gas that is produced.Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
agasfer [191]3 years ago
4 0
Auuuuuhhhhhhh auhhh Frog
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A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in different proportions. Which of the followi
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Answer:

The information that can be used to determine which mixture has the higher proportion of KCl IS INFORMATION ABOUT THE MASS OF CHLORINE IN EACH MIXTURE, THIS INFORMATION CAN BE OBTAINED BY USING THE LAW OF DEFINITE PROPORTION.

Explanation:

The law of definite proportion states that the chemical composition by mass of a chemical compound is always constant. For instance, a chemical compound that is made up of two elements will always contain the same proportions of the constituent elements regardless of the quantity of chemical that was used.

Using the law of definite proportion, we can determine the proportion of sodium and chlorine in NaCl and the proportion of potassium and chlorine in KCl if the mass of chlorine that was used is known. Based on the results obtained, one can easily determine the mixtures that has higher proportion of KCl.

6 0
3 years ago
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What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon
Varvara68 [4.7K]

Answer: 31.8 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles

\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles

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According to stoichiometry :

1 mole of Al_2O_3 require 3 moles of C

Thus 0.59 moles of Al_2O_3 will require=\frac{3}{1}\times 0.59=1.77moles  of C

Thus Al_2O_3 is the limiting reagent as it limits the formation of product and C is the excess reagent as it is present in more amount than required.

As 1 mole of Al_2O_3 give = 2 moles of Al

Thus 0.59 moles of Al_2O_3 give =\frac{2}{1}\times 0.59=1.18moles  of Al

Mass of Al=moles\times {\text {Molar mass}}=1.18moles\times 27g/mol=31.8g

Thus 31.8 g of Al will be produced from the given masses of both reactants.

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Licemer1 [7]

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