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OLga [1]
2 years ago
9

Using complete sentences, explain how to predict the products and balance the reaction for the combustion of butane.

Chemistry
1 answer:
Tanzania [10]2 years ago
8 0

Answer:

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H2₀

<h2>What is combustion? </h2>

The reaction of a substance with oxygen, producing one or more oxides, heat and light.

Butane (C₄H₁₀) is a hydrocarbon ( The name used for a hydrocarbon with four carbon atoms and only single bonds.) Butane consists of carbon and hydrogen and thus when it combust, it produces carbon dioxide and water.

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H2₀

The following chemical equation follow the law of conservation of mass.

The law of conservation of mass states that in a closed system, mass is neither created nor destroyed during a chemical or physical reaction. The law of conservation of mass is applied whenever you balance a <u>chemical equation</u>. It is also applicable to stoichiometry. On the product side, the number of atoms must match the total of atoms on the product side. As a result, the equation must be balanced.  

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Advantages and disadvantages of titration?
alisha [4.7K]

Answer:

What are the advantages of titration?

Titrimetric analysis commonly referred to as volumetric analysis offers distinct advantages over cumbersome gravimetric methods:

Speed of analysis.

Instantaneous completion of reactions.

Greater accuracy due to minimization of material loss involved in decanting, filtration, precipitation or similar operations.

Explanation:

Disadvantages

It is a destructive method often using up relatively large quantities of the substance being analysed.

It requires reactions to occur in a liquid phase, often the chemistry of interest will make this inappropriate.

It can produce significant amounts of chemical waste which has to be disposed of.

It has limited accuracy.

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5 0
3 years ago
An atom that has a full valence electron shell can be found in which group?
kaheart [24]

Answer:

Noble Gasses

Explanation:

4 0
2 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
wolverine [178]

Answer:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)

∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)

∴ O /  Al: 0 → +2 ≡ 2e-

         Na: +1 → +2

∴ R /  H: +1 → 0

     2 - Al - 2

     2 - Na - 1

     8 - O - 8

     14 - H - 14

⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

           1 - S - 1

           4 - O - 4

           2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

       8 - P - 8

       12 - O - 12

⇒ Kc = 1 / PO2∧6

6 0
2 years ago
How many grams of aluminum oxide will you need to start with in order to make 6500 g of aluminum hydroxide?
kherson [118]

Answer:

Explanation:

2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.

3 0
2 years ago
Read 2 more answers
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