I think it's call a magnet
Answer:
D
Explanation:
I believe it is D. your kinetic energy would be at b. A, the cart would be going at a constant rate, because there is no hill or steep slope.
<h3>Answers:</h3>
1) 2 Units of Ozone
2) 3 Units of Ozone
3) 9 Units of Ozone
<h3>Solution:</h3>
1) From 6 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
6 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 2 Units of Ozone
2) From 9 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
9 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 3 Units of Ozone
3) From 27 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
27 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 9 Units of Ozone
Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
