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marin [14]
3 years ago
12

Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

he following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O 16 Δ[H2O]Δt 14 Δ[NO2]Δt - 14 Δ[NH3]Δt - 17 Δ[O2]Δt All of the above are valid expressions of the reaction rate.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

4NH_3+7O_2\rightarrow 4NO_2+6H_2O

The expression for rate of reaction for the reactant :

\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}

\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}

The expression for rate of reaction for the product :

\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}

From this we conclude that, all the options are correct.

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Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b
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Answer:

Explanation:

The cell reaction properly written is shown below:

              Cu|Cu²⁺_{aq} || Ag⁺_{aq} | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

  Oxidation half:

                  Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

At the anode, oxidation occurs.

  Reduction half:

                  Ag⁺_{aq} + 2e⁻ ⇄ Ag_{s}

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

             Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

              Ag⁺_{aq} + e⁻ ⇄ Ag_{s}

  we multiply the second reaction by 2 to balance up:

         2Ag⁺_{aq} + 2e⁻ ⇄ 2Ag_{s}

The net reaction equation:

Cu_{s} + 2Ag⁺_{aq} + 2e⁻⇄ Cu²⁺_{aq} + 2e⁻ + 2Ag_{s}

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

  Cu_{s} + 2Ag⁺_{aq} ⇄ Cu²⁺_{aq} + 2Ag_{s}

6 0
3 years ago
How many moles of atoms are in 7.00 g of 13c? express your answer numerically in moles?
Nadusha1986 [10]
Molar mass of 13c = 13 grams
number of moles = mass / molar mass
therefore,
number of moles = 7 / 13
To know the number of atoms in 7/13 moles, we simply multiply the number of moles by Avogadro's number as follows:
number of atoms = (7/13) x 6.022 x 10^23 = 3.2426 x 10^23 atoms
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Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production
Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>

<em></em>

Thus <em>ΔH° of the reaction is -747.54kJ</em>

3 0
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