Answer:
oxygen is responsible for rusting
Formic acid when in water would dissociate into ions just like any acids. It would dissociate into the hydrogen ion and the formate ion. The equilibrium dissociation equation would be written as:
<span>HCOOH (aq) + H2O (l) ⇌ H+ (aq) + HCOO- (aq)
Formic acid is a weak acid which means that when in aqueous solution it does not completely dissociate into its corresponding ions. Only a certain amount that would be dissociated so in the solution there will be HCOOH, HCOO- and H+ molecules. It is also known as Methanoic acid and an important substance for the synthesis of a number of substances. It is naturally occurring in ants.</span>
The correct is D.
Water is a polar molecule and it has polar bonds, which carry partially positive and partially negative charges. This polar bond increases the attraction between molecules of water and thus it requires a greater energy to break the bond between the molecules of water compare to carbon dioxide, which is a non polar molecule. Thus, water has a higher boiling point than carbon dioxide.
Answer:
The chemistry of iron is dominated by the +2 and +3 oxidation states i.e. iron(II) and iron(III) complexes e.g. Fe2+ and Fe3+ complex ions with selected ligands, usually of an octahedral shape, a few tetrahedral iron(III) complexes are mentioned too. The reactions of the aqueous ions iron(II) and iron(III) with ammonia, sodium hydroxide and sodium carbonate are described and explained as are complexes of iron(III) with the chloride ion and cyanide ion.
principal oxidation states of iron, redox reactions of iron, ligand substitution displacement reactions of iron, balanced equations of iron chemistry, formula of iron complex ions, shapes colours of iron complexes, formula of compoundsExplanation:
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
