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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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A patient is required to take an IV drug for cancer treatment. The required dosage is 5.0 mg drug her lb patient body weight eve
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3 bags are required.

Explanation:

Find the number of mg needed for 150 pounds

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and
rusak2 [61]

Answer:

4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)

-) Oxidizing agent: NO_3^-_(_a_q_)

-) Reducing agent: Zn_(_s_)

Explanation:

The first step is separate the reaction into the <u>semireactions</u>:

A.Zn~->Zn(OH)_4^-^2

B.NO_3^-~->~NH_3

If we want to balance in <u>basic medium </u>we have to follow the rules:

1. We adjust the oxygen with OH^-

2. We adjust the H with H_2O

3. We adjust the charge with e^-

Lets balance the first semireaction A. :

Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-

Now, lets balance semireaction B:

NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-

Finally, we have to add the two semireactions:

_________________________________________

8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)

2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)

_________________________________________

(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)

(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)

Cancel out the species on both sides:

8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3

Simplifying the equation :

4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3

The Zn_(_s_) is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The NO_3^-_(_a_q_)is<u> reduced</u> therefore is the <u>oxidizing agent</u>.

4 0
4 years ago
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