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Triss [41]
3 years ago
6

Write a balanced equation for the complete oxidation reaction that occurs when glucose (C6H12O6) reacts with oxygen. Use the sma

llest possible integer coefficients.
Chemistry
1 answer:
Stels [109]3 years ago
7 0

Answer:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Explanation:

Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).

The equation can be written as follow:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

The above equation can be balance as illustrated below:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O

There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Now, the equation is balanced.

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3 years ago
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Describing Metallic Bonding Theories
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5 0
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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

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3 years ago
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