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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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daser333 [38]

Answer:

6 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v          (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = <em>0.05 L</em>

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of H and O.

H= 1.01 g/mol         O= 16.00 g/mol

2.) Multiply them and then add them together to have their combined molecular weights. (You have to multiply by 2 because of their equation; H2O2).

2(1.01) + 2(16.00)= 34.02 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 34.02 g = <em>0.2939 mol</em>

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4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.2939 mol / 0.05 L = <em>5.878 mol/L</em>

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 5.878 to 6 mol/L

Sorry this explanation is very long let me know if you need a better more written out explanation.

4 0
3 years ago
He diagram below represents part of the oxygen cycle.
salantis [7]

Answer:

The answer is C

Explanation:

7 0
3 years ago
How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only
forsale [732]

Answer:

m_{Fe_2O_3}=549gFe_2O_3

Explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

4Fe+3O_2\rightarrow 2Fe_2O_3

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3}  \\\\m_{Fe_2O_3}=549gFe_2O_3

Regards!

8 0
3 years ago
What is the mole fraction of ethanol in a solution of 3.00 moles of ethanol and 5.00 moles of water?
Rus_ich [418]
If theres a mixture of components we can calculate the mole fraction
mole fraction can be calculated as follows
mole fraction of component = \frac{number of moles of component}{total number of moles of all components }
number of moles of ethanol - 3.00 mol
total number of moles in mixture - 3.00 + 5.00 = 8.00 mol 

mole fraction of ethanol = \frac{3.00 mol}{8.00 mol }
mole fraction of ethanol is 0.375
6 0
3 years ago
Complete the mechanism for the acid-catalyzed hydrolysis of the epoxide in alcohol by adding any missing atoms, bonds, charges,
Sauron [17]

Answer:

See explanation below

Explanation:

You are missing the structure, therefore, I will do an example with one that I found on another place to try to explain.

This acid mechanism always involves carbocations, and positive charges, never negative because we are in acidic mediums.

In the first step, the lone pairs of the oxigen from the epoxide, substract one hydrogen of the reactant.

Second step, the lone pairs of the oxygen from the reactant, do a nucleophylic attack to the carbon of the epoxide. In this case, it will do it to the most substitued carbon.

Then, in the third step by acid base equilibrium, the hydrogen from the reactant that attacked, is substracted from the molecule by a molecule of water (We are in acid medium, therefore, there is traces of water) and the final structure is formed.

Check picture for mechanism:

4 0
3 years ago
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