Fluoride is an anion of Fluorine
What this means is that the two have the same number of protons (9), but Fluoride has 10 electrons compared to Fluorine's 9.
So the answers are:
Protons - 9
Neutrons - 9
Electrons - 10
Atomic Number - 9
Atomic Mass - 19 g/mol
Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Molarity = moles of solute(HCl)
------------------------------------
volume of the solution
= 1
------
5
= 0.2M.
Hence option B is correct.
Hope this helps!!
When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
