<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 and Explanation:
For the following balanced reaction:
PCl₅(g) ↔ PCl₃(g) + Cl₂(g)
We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:
Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
Define stoichiometry and explain why it is such a useful tool for chemists. ...Explain<span> what a limiting reactant is and why it is important in </span>stoichiometry<span>. A limiting reactant is the reactant in a chemical equation that runs out first. It is important because its quantity determines the amount of products produce</span>
Answer:
D. 4/2 He corresponds to an alpha particle.
