Answer:
Part A: 2N₂O(g) ⇄ 2N₂(g) + O₂(g)
Part B: -r = K*[N₂O]²
Part C: K= k1*k2
Explanation:
Part A
To do the balance chemical question for the overall chemical reaction, we must sum the reaction of the steps, eliminating the intermediaries, which are the compounds that have the same amount both at reactants and products (bolded).
N₂O(g) ⇄ N₂(g) + O(g)
N₂O(g) + O(g) ⇄ N₂(g) + O₂(g)
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2N₂O(g) + O(g) ⇄ 2N₂(g) + O(g) + O₂(g)
2N₂O(g) ⇄ 2N₂(g) + O₂(g)
Part B
The velocity of the reaction (r) can be calculated based on the reactants or based on the products. Let's do it based on the disappearing of the reactant. Because it is disappearing, the variation at its concentration must be negative, so the rate will be negative.
Let's suppose its an elementary reaction, so, the concentration of the reactant must be elevated by its coefficient. And let's call the overall rate constant as K:
-r = K*[N₂O]²
Part C
Because the steps were summed, and the reactions were not multiplied by a constant or inverted, the constant K is just the multiplication of the constants of the steps:
K= k1*k2
Photosynthesis: The process of making something ( carbohydrates ) with light
CO+H2O &Sunlight=O2+C6H6O12
The answer is Straight Edge.
<h3><u>Answer;</u></h3>
increase the partial pressure of CO2
<h3><u>Explanation;</u></h3>
2CO(g) + O2(g) -------> 2CO2(g) + heat
- Removel of Oxygen makes the equilibrium to move towards the left side (reactants) because the reactants concentration is decreasing. According to the LeChâtelier’s principle equilibrium moves towards the lower concentration.
- When the partial pressure of any of the gaseous reactants or of the products is increased, the position of equilibrium is shifted so as to decrease its partial pressure. In this case, the removal of oxygen causes a shift to to the left which causes an increase in the partial pressure of CO2.
Answer:
three spaces to the left
Explanation:
this because there are 1000 meters in every kilometer
