Explanation:
NiCl2 + KOH = NiOH + KCl
since nickel is the primary element in nickel ii chloride, find the oxidation no of nickel in the compound and chlorine should have 2 as a subscription since the ii in the compound represents that nickel has an oxidation no of 2
1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.
2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.
3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.
4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
We have that the work function of the metal
From the Question we are told that
UV radiation (λ = 162 nm)
Kinetic energy 
Generally the equation for Kinetic energy is mathematically given as
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Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
