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4vir4ik [10]
2 years ago
13

Methane (CH4), is said to be the cleanest burning fossil fuel. When burned, it releases carbon dioxide gas and forms water. Usin

g the equation below, determine which of the following is true (assuming there is an infinite supply of oxygen gas (O2)). Select all that apply.
For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.


For every 2 moles of methane (CH4) that reacts, 2 moles of H2O are produced.


For every 2,000 molecules of methane (CH4) that reacts, 2,000 molecules of H2O are produced.


For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.


For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.
Chemistry
1 answer:
LiRa [457]2 years ago
3 0

From the stoichiometry of the balanced reaction equation, the correct statement are;

  • For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
  • For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
  • For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.

<h3>What is combustion?</h3>

The term combustion refers to the burning of fossil fuels for the purpose of energy production. The equation for reaction is CH4 + 2O2 ---> CO2 + 2H2O.

Using this equation as shown, the true statements are;

  • For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
  • For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
  • For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.

Learn more about combustion: brainly.com/question/15117038

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The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

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Answer:

Explanation:

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Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO  </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em>  will produce NH₄⁺. 0.25mol HCl consume all NH₃.

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Explanation:

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