What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 195.0 ml of a 0.194 m lead(ii) nitrate solution?

First, we write the reaction equation:

2KI + PbNO₃  → K₂NO₃ + PbI₂
The molar ratio of KI to PbNO₃ is 2 : 1
Moles of PbNO₃ present:
Moles = concentration (M) x volume (dm³)
= 0.194 x 0.195
= 0.038
Moles of KI required = 2 x 0.038 = 0.076 moles
concentration = moles / volume
volume = moles / concentration
= 0.076 / 0.2
= 0.38 L = 380 ml

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.

                             

What do you mean by that? like the contents? the thc level?