Why are the elements 57 through 71 and 89 through 102, found separately at the bottom of the table?

1 answer:

8 0

Elements from 57 to 71 are Lanthanoids and from 89 to 103 are Actinoids. Simply I would say Actinoids have less similarity in chemical properties than Lanthanoids. On the other hands, Actinoids have randomized chemical strutural bonds.

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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /

barxatty [35]

Answer:

$[Pb^{2+}]=3.9 \times 10^{-2}M$

this is the concentration required to initiate precipitation

Explanation:

$PbCl_2$ ⇄ $Pb^{2+}+2Cl^-$

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

$Ip=[Pb^{2}][2Cl^-]^2=Ksp$

$Ksp=2.4\times 10^{-4}$

lets solubility=S

$[Pb^{2+}] = S$

$[Cl^-]=2S$

$Ksp=[Pb^{2+}]\times [Cl^-]^2$

$Ksp=S \times (2S)^2$

$Ksp=4S^3$

$S=\sqrt[3]{\frac{Ksp}{4} }$

$S=3.9\times 10^{-2}$

$[Pb^{2+}]=3.9 \times 10^{-2}M$ this is the concentration required to initiate precipitation

4 0

3 years ago

Describe an acid or base in terms of its relationship to protons

motikmotik

Acids are donors of protons (H+) and bases are acceptors of protons.
For example:
1) hydrochloric acid (HCl) in reaction with water give one proton to water and become chloride anion (Cl-).
2) ammonia (NH3) Is base, in reaction with water accepts one protone and become ammonium cation (NH4+).

4 0

3 years ago

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m

gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$