How many moles of Ca(NO3)2 must be added to 1.0 L of a 0.255 M HF solution to begin precipitation of CaF2(s)

6.1×10^(-10) mol

Explanation:

CaF2(s) > Ca2+(aq) +2F-(aq)

          Ksp = [Ca2+][F-]^2

  Given data

   [F-] = 0.255 M

∴ The concentration of Ca2+required to start the precipitation

                 [Ca2+] = Ksp / [F-]2

                            = (4.0 X 10^-11) /(0.255)^2

                            = 6.1×10^(-10) M

Here the volume is 1.0 L  

∴ The number of molesof  Ca(NO3)2 must be added to 1.0 L of a1.0 L of a 0.255 M HF solution to begin precipitation ofCaF2 (s)=   6.1×10^(-10) mol

the answer would be a

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.