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QveST [7]
3 years ago
12

The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.

5 kJ/mol; and H2 O(g): Hf = –241.82 kJ/mol. How much heat is released by the combustion of 2 mol of methane? Use .
Chemistry
2 answers:
algol [13]3 years ago
8 0

Answer:

The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol

Explanation:

The chemical reaction of the combustion of methane is given as follows;

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor

Where:

CH₄ (g): Hf = -74.6 kJ/mol

CO₂ (g): Hf = -393.5 kJ/mol

H₂O (g): Hf = -241.82 kJ/mol

Therefore, the combustion of 1 mole of methane releases;

-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol

Hence the combustion of 2 moles of methae will rellease;

2 × -802.54 kJ/mol or 1,605.08 kJ/mol.

vagabundo [1.1K]3 years ago
4 0

Answer: answer is C

Explanation:

Just took the test

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Which of the following conclusions might a scientist make when fossils of two different species of very similar organisms are fo
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Answer:

if its multiple, C and D, if not, C

Explanation:

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3 years ago
If 125g of KClO3 is heated, what is the total mass of the products?​
Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

  1. Find the number of moles of the given reactant:

     Number of moles  = \frac{mass}{molar mass}

  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

Mass  = number of moles x molar mass

  molar mass of KCl  = 39 + 35.5 = 74.5g/mol

  molar mass of O₂  = 16 x 2  = 32g/mol

  Mass of KCl  = 74.5 x 1.02  = 75.99g

  Mass of O₂  = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

                                        = 124.95g

This value is approximately the same as that of mass of  KClO₃

 

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3 years ago
Your body is much larger now than it was when you were a baby. What is the cause of this growth? *
masya89 [10]

Answer:c

Explanation:

4 0
3 years ago
Read 2 more answers
600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c
Ludmilka [50]

Answer:

k=1.20x10^{-3} s^{-1}

Explanation:

For a first order reaction the rate law is:

v=\frac{-d[A]}{[A]}=k[A]

Integranting both sides of the equation we get:

\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

ln\frac{[A]}{[A]_{0} } =-kt

[A]=[A]_{0}e^{-kt}

ln[A]=ln[A]_{0} -kt

k=\frac{ln[A]_{0}-ln[A]}{t}

therefore k is

k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}

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4 years ago
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I think that type of rock is call Igneous rock because the igneous rock is molten rock from volcanoes. 
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