The ratio of the diffusion rate of Cl2 and O2 is 1.5
calculation
rate of diffusion Cl2/ rate of diffusion O2 =√ molar mass of Cl2/ molar mass of O2
molar mass of Cl2 = 35.5 x2 = 71 g/mol
molar mass O2 = 16 x2 =32g/mol
that is rate of diffusion Cl2/ rate of diffusion of O2 =√ 71/ 32= 1.5
The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</span>
If there are options, they are b, c, and d. I've done this question in class before :)
Answer: pH=12.69
Explanation:



Initial 0.12 0 0
Eqm 0.12-x x x
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
(neglecting small value of x in comparison to 0.12)

Moles of 



0.06 moles of NaOH will give 0.06 moles of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
Now
moles of
will be neutralized by
moles of
and
moles of
will be left.
Molarity of 
![pOH=-\log[OH^-]=-\log[0.049]=1.31](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.049%5D%3D1.31)
pH = 14 - pOH= 14 - 1.31 = 12.69
w/w percentage <span>
= mass of the pure compound /
total mass of the sample x 100%
70% HNO₃
contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>
The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
= 103.8 g x (70 / 100)
<span>
= 72.66 g = 72.7 g.</span></span>