Answer:- The hydroxide ion concentration of the solution is
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
= 0.00000195 or 
So, the hydroxide ion concentration of the solution is
.
Answer:
volume in L = 0.25 L
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 2.43 g
Volume of KI = ?
Solution:
Balanced chemical equation:
2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 2.43 g/ 187.56 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of Cu(NO₃)₂ with KI.
Cu(NO₃)₂ : KI
2 : 4
0.013 : 4 × 0.013=0.052 mol
Volume of KI:
<em>Molarity = moles of solute / volume in L</em>
volume in L = moles of solute /Molarity
volume in L = 0.052 mol / 0.209 mol/L
volume in L = 0.25 L
Answer:
See the figure
Explanation:
In this case, we have to take into account the <u>stability of the carbocations</u>:
Terciary>Secundary>>Primary.
In other words, is we have the <u>most substituted carbocation</u> we will have more stability. Therefore in the carbocation formation, the charge would go in the <u>most substituted carbon</u> of the double bond for each case.
Barium, it's an earthy alkaline metal it's atomic number is 56