<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:
Following laboratory safety guidelines minimizes the chance of lab accidents.
Explanation:
Answer:
(a) ![pH=1](https://tex.z-dn.net/?f=pH%3D1)
(b) ![pH=1.3](https://tex.z-dn.net/?f=pH%3D1.3)
(c) ![pH=13](https://tex.z-dn.net/?f=pH%3D13)
(d) ![pH=12.7](https://tex.z-dn.net/?f=pH%3D12.7)
Explanation:
Hello,
In this case, we define the pH in terms of the concentration of hydronium ions as:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:
(a)
![[H^+]=[HCl]=0.1M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D0.1M)
(b)
![[H^+]=[HCl]=0.05M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D0.05M)
![pH=-log(0.05)=1.3](https://tex.z-dn.net/?f=pH%3D-log%280.05%29%3D1.3)
Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:
![[OH^-]=[NaOH]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BNaOH%5D)
![pOH=-log([OH^-])](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29)
![pH=14-pOH](https://tex.z-dn.net/?f=pH%3D14-pOH)
Thus, we have:
(b)
![pOH=-log(0.1)=1\\pH=14-1=13](https://tex.z-dn.net/?f=pOH%3D-log%280.1%29%3D1%5C%5CpH%3D14-1%3D13)
(d)
![pOH=-log(0.05)=1.3\\pH=14-1.3=12.7](https://tex.z-dn.net/?f=pOH%3D-log%280.05%29%3D1.3%5C%5CpH%3D14-1.3%3D12.7)
Best regards.