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trasher [3.6K]
3 years ago
10

During the combustion of methane in water, which bonds are broken and which bonds are formed? Sort the molecules based on whethe

r their bonds are broken or formed during the combustion reaction.
Chemistry
1 answer:
melomori [17]3 years ago
8 0

Answer:

The bonds which are broken are:-

Bonds of C-H in the CH_4

Bonds of O-O in the O_2

Bonds which are formed are:-

Bonds of C-O in the CO_2

Bonds of O-H in the H_2O

Explanation:

The combustion reaction of methane is shown below as;-

CH_4+2O_2\rightarrow CO_2+2H_2O

Covalent bond is the bond which is formed with the sharing of the electrons between the two atoms which are taking part in the bond. It is generally formed between the atoms with similar electronegativity values. All the bonds in the above equation are covalent bonds.

The bonds which are broken are:-

Bonds of C-H in the CH_4

Bonds of O-O in the O_2

Bonds which are formed are:-

Bonds of C-O in the CO_2

Bonds of O-H in the H_2O

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What is an octet of electrons? Which elements contain an octet of electrons?
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When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu
Gwar [14]

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

<em>1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄</em>

<em />

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

<em>Moles S₈: 256.52g/mol</em>

10.0g * (1mol / 256.52g) = 0.0390 moles

<em>Moles F₂:</em>

<em>PV = nRT</em>

PV/RT = n

<em>Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)</em>

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

6 0
3 years ago
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