The products of a combustion reaction include:

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O + energy In the reaction listed above, 1 molecule of glucose reacts with 6 molecules of oxyg

kirill115 [55]

Answer:

Third choice: energy present in the glucose and oxygen that is not needed for the formation of carbon dioxide and water is released to form energy/ATP.

Explanation:

1) Chemical equation (given):

C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

2) Chemical potential energy:

Each compound stores chemical potential energy. This energy is stored in the chemical bonds.

Due to every substance has its own unique chemical potential energy, when a chemical reaction takes plase, yielding to the change of some substances, some energy is absorbed (when bonds are formed) and some energy is released (when bonds are broken).

3) Conservation of energy:

Then, if the sum of the bond energies of the final products is less than the sum of the bond energies of the reactants, the law of conservation of energy rules that the difference between the total energies of the products and reactants must be released to the surroundings.

That is what is happening in the given reaction:

C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

The term energy in the product side means that energy is conserved because it is being released due to the the glucose and oxygen (reactant side) have more energy stored in their bonds than the energy needed for the formation of carbon dioxide and water, so that excess of energy is released to form energy/ATP.

Summarizing:

The energy on the product side added to the energy of carbon dioxide and water equals the energy of the glucose and oxygen and the final balance is:

∑ Energy of the reactants = ∑energy of the products + released energy, supporting the law of conservation of energy.

5 0

4 years ago

Give the charge on the transition metal of each of the following compounds:

elena-s [515]

Au +2
Fe +2
Zn+2
Here’s my periodic table. I hope it helps

7 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

(c) Δ $S_{sys}$ = 0 ; Δ $S_{sur}$ = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$