Answer:
NH3 weak base
HCOOH weak acid
H3PO4 weak acid
NaOH strong base
Explanation:
When talking about the weakness or strength of acid and bases, it is essential to look at it from the base point of ionization or dissociation in solutions. For an acid or base to be termed strong , this means it ionize completely in solution to form hydroxonium ion and hydroxide jobs respectively in water. For an acid or base to be termed weak, it ionizes only partially in solution.
The strength of acid can be accessed from the value of the acid ionization constant Ka. Strong acids like sulphuric acid have a fairly large value for for this constant while weak acids have a relatively low value for this constant. Hence, we must know that the dissociation equation is not enough to say an acid is weak or strong. Acids with higher Ka value are stronger with acids with lower Ka value.
In likewise manner, we have the base dissociation constant. It is the equilibrium constant for a base dissociation reaction. While stronger bases have relatively higher values for Ka, weak bases have a relatively lower value.
Ammonia yields a weak base solution of ammonium hydroxide in water. This makes it a weak acid
Methanoic or Formic acid like most organic acid is a weak acid. It ionizes only partially in solutions.
Phosphoric acid is also a weak acid as it ionizes partially in solutions too.
Lastly, sodium hydroxide is a strong base as it ionizes completely in solution
Answer:
ΔH°C2H2Cl4(l) = -333,36 kJ/mol
ΔH°r₂ = -35,14 kJ/mol
Explanation:
The ΔH°r of the first reaction is:
ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))
ΔH°H2(g) = 0 kJ/mol
ΔH°C2H4(g) = 52,4 kJ/mol
Δ°HCl2 (g) = 0 kJ/mol
Replacing:
ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>
The standard heat of the second reaction is:
ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)
Where:
ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol
Replacing:
ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol
<em>ΔH°r₂ = -35,14 kJ/mol</em>
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I hope it helps!
MnO2 is the formula for manganese (4) dioxide
The complex ion has a charge of -3 which means it is an anionic complex. The nomenclature for this complex can be broken in to two parts. The part that contains the cyano and the part that contains aluminium.
Cyano Ligand
There are 6 cyanide ions in the complex ion and so we assign the prefix hexa to explain the presence of 6 anions. This makes our ligand part be name hexacyano.
The aluminium part
Since the metal ion ends up in an anionic complex we add -ate at the end of aluminium to show that we have an anionic complex.
Henceforth we combine the two parts and start with the ligand in naming the complex. is called hexacyanoaluminate