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zhenek [66]
2 years ago
13

Classify the outcomes based on whether they are caused by adding or removing a reactant from a chemical reaction. The rate of th

e forward reaction increases. The rate of the reverse reaction increases. The concentration of products increases. The concentration of products decreases.
Chemistry
1 answer:
Mama L [17]2 years ago
3 0

Explanation: To study the outcomes, we will apply Le-Chatelier's principle.

Le-Chatelier's principle states that if there is any disturbance in the conditions of the dynamic equilibrium, the position of equilibrium will counteract the change.

1) The rate of the forward reaction increases.

This will happen when we add the reactant to a chemical reaction. According to Le-Chatelier's principle, by increasing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, forward reaction is favored.

2) The rate of the revere reaction increases.

This will happen when we remove the reactant from a chemical reaction. According to Le-Chatelier's principle, by removing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, reverse reaction is favored.

3) The concentration of product increases.

This will happen when we add reactants to a chemical reaction. According to  Le-Chatelier's principle, when we increase the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the forward direction which means that the concentration of product will increase.

4) The concentration of products decreases.

This will happen when we remove reactants from a chemical reaction. According to  Le-Chatelier's principle, when we decrease the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the reverse direction which means that the concentration of reactants will increase or concentration of products will decrease.

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The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

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                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

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                        x = 0.891 g of Carbon

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                       0.891 g of C     ----------   x

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2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

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Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

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Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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