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Setler [38]
3 years ago
10

list 10 chemical reactions that have benefited your life today. Include the reasons you think each was indeed a chemical reactio

n and not just a physical change. Thank about processes in your body, processes in the atmosphere, or chemical reaction later that may be involved with any materials or products you have used.
Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
4 0

1) The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

Plants convert solar energy into the chemical energy of sugars (food).

2) Chemical reaction of cellular respiration:  

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Organic carbon (glucose) is transformed into inorganic carbon (carbon dioxide) that goes into atmosphere.

3) Gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O.

Gasoline is a mixture of many different hydrocarbons: alkanes (paraffins), cycloalkanes and alkenes (olefins).

Gasoline is being burn.

4) Balanced chemical reaction of forming rust: 4Fe + 3O₂ → 2Fe₂O₃.

Forming rust usually last for few months and iron changing in another substance with different properties.

5) An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.  

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).  

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

Copper is forming from the solution, that is chemical change.

6) Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.

Making limewater  (diluted solution of calcium hydroxide).

7) Neutralization is reaction in which an acid and a base react quantitatively with each other. In this reaction, there are no acids and bases, only salts.

Balanced chemical reaction of vinegar and antacid:

2CH₃COOH(aq) + Mg(OH)₂(aq) → (CH₃COO)₂Mg(s) + 2H₂O(l).

Precipation is formed.

8) Balanced chemical reaction of magnesium and hydrochloric acid:

MgO + 2HCl → MgCl₂ + H₂O(g).

Acid dissolves metal oxides.

9) Chemical reaction of silver in air:

4Ag + 2H₂S + O₂ → 2Ag₂S (tarnish) + 2H₂O.

Silver change color in the air.

10) Forming of patina: 2CuO + CO₂ + H₂O → Cu₂CO₃(OH)₂.

Brass (copper) change color to green.

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Given the following data:
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176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
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Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

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