the formula we is as follows:-
M1V1= M2V2
where
M1=1.2
V1=0.133l
V2=41l
M2=?
1.2 × 0.133 = 41 × M2
0.1596 = 41 × M2
M2 = 0.15960/41
M2 = 0.0038926829
Answer:
Explanation:
MM of I2 = 2 (127 g) = 254 g/mol
0.065 mol I2 x 254g I₂/ 1 mol I₂ = 16.5 g I2
<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
<u>Explanation:</u>
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%282%5Ctimes%200.00066%29%3D1.32%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above equation, we get:

Hence, the hydroxide ion concentration and pOH of the solution is
and 2.88 respectively