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Fantom [35]
2 years ago
14

When ethanol (C2H6O(aq)) is consumed, it reacts with oxygen gas (O2) in the body to produce gaseous carbon dioxide and liquid wa

ter. Enter the balanced chemical equation for the reaction
Chemistry
2 answers:
ankoles [38]2 years ago
6 0

Answer:

C2H6O(aq) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation:

Step 1: data given

The reactants are:

ethanol = C2H6O(aq)

oxygen gas = O2(g)

The products are:

gaseous carbon dioxide = CO2(g)

liquid water = H2O(l)

Step 2: The unbalanced equation

C2H6O(aq) + O2(g) → CO2(g) + H2O(l)

Step 3: Balancing the equation

C2H6O(aq) + O2(g) → CO2(g) + H2O(l)

On the left side we have 2 time C (in C2H6O), on the right side we have 1x C (in CO2). To balance the amount of C on both sides, we have to multiply CO2 on the right side by 2.

C2H6O(aq) + O2(g) → 2CO2(g) + H2O(l)

On the left side we have 6x H (in C2H6O), on the right side we have 2x H(in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 3.

C2H6O(aq) + O2(g) → 2CO2(g) + 3H2O(l)

On the right side we have 7x O (4x O in 2CO2 and 3x O in 3H2O), on the left side we have 3x O (1x in C2H6O and 2x in O2). To balance the amount of O on both sides, we have to multiply O2 on the left side by 3. Now the equation is balanced.

C2H6O(aq) + 3O2(g) → 2CO2(g) + 3H2O(l)

jonny [76]2 years ago
5 0

Answer:

C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O

Explanation:

Ethanol (C₂H₆O) is processed by the body when reacts with oxygen (O₂) producing carbon dioxide (CO₂) and water (H₂O). The unbalanced reaction is:

C₂H₆O + O₂ → CO₂ + H₂O

As you can see, ethanol contains 2 atoms of carbon, that means must be produced 2 CO₂:

C₂H₆O + O₂ → <em>2</em> CO₂ + H₂O

Also, ethanol contains 6 atoms of hydrogen that produced 3 molecules of water:

C₂H₆O + O₂ → <em>2</em> CO₂ + <em>3</em> H₂O

Now, with carbon and hydrogen balanced, there are produced 7 atoms of oxygen. As ethanol contains 1 atom of oxygen. 3 moles of O₂ must react in the beginning. Thus, balanced reaction is:

C₂H₆O + <em>3</em> O₂ → <em>2</em> CO₂ + <em>3</em> H₂O

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What is the value of Keq for the reaction expressed in scientific notation? 2.1 x 10-2 2.1 x 102 1.2 x 103 1.2 x 10-3
Lady_Fox [76]

Complete question:

Consider the reaction.

At equilibrium at 600 K, the concentrations are as follows.

2HF -----> H₂ + F₂

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3 M

[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?

2.1 x 10-2

2.1 x 102

1.2 x 103

1.2 x 10-3

Answer:

2.1 × 10^-2

Explanation:

Kequilibrum(Keq) = product/reactant

Equation for the reaction :

2HF -----> H₂ + F₂

Therefore,

Keq = [H2][F2] / [HF]^2

Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2

Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]

Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]

Keq = 2.0665 × 10^(-6 - (-4))

Keq = 2.0665 × 10^(-6 + 4)

Keq = 2.1 × 10^-2

7 0
2 years ago
Read 2 more answers
Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca
lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.23-0.77=-1.0V

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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2 years ago
Jill is doing an experiment on the movement of pill bugs. She will place the pill bugs on flat surfaces covered with diffirent m
Ostrovityanka [42]

Answer:D

Explanation:

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A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is
Novosadov [1.4K]

Answer:

8.09x10⁻⁵M of Fe³⁺

Explanation:

Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.

In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-

The concentration of Fe³⁺ in the reference sample is:

4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺

<em>Because reference sample was diluted from 5.0mL to 50.0mL.</em>

<em>That means a solution of  4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512</em>

Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:

0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = <em>3.23x10⁻⁵M.  </em>

As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:

3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = <em>8.09x10⁻⁵M of Fe³⁺</em>

4 0
2 years ago
This means that the steel bar lost 14900 J of thermal energy. What is the change in temperature of the steel bar? Recall that th
stiv31 [10]

Answer:

ΔT=-747,13°C

Explanation:

Sensible heat is<em> the amount of thermal energy that is required to change the temperature of an object</em>, the equation for calculating the heat change is  given by:

Q=msΔT

where:

  • Q, heat that has been absorbed or realeased by the substance [J]
  • m, mass of the substance [g]
  • s, specific heat capacity [J/g°C] (
  • ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

ΔT=\frac{-14900 J}{40,7g*0,49\frac{J}{gC} }=-747,13°C

<em>(Note that Q=-14900 J because there is a </em><u><em>LOST</em></u><em> of thermal energy)</em>

Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.

5 0
2 years ago
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