Answer:
Explanation:
If this is a double replacement reaction or salt metathesis, then I have taken the liberty of including the symbol representing an aqueous solution, (aq), to the formulas in the equation written as CuCl2(aq)+2NaNO3(aq)→Cu(NO3)2(aq)+2NaCl(aq).
In order for a double replacement reaction to occur, one of the products must be a solid precipitate (a salt), an insoluble gas, or water. Since all species are in aqueous solution, none of these products occur, which means there is no reaction. The result is an aqueous solution containing Cu2+, Cl−, Na+, and NO3− ions.
The equation should be written as CuCl2(aq)+2NaNO3(aq)→No Reaction
2. CuCl₂·2H₂O is limiting reactant.
Chemical reaction: 3CuCl₂·2H₂O + 2Al → 3Cu + 2AlCl₃ + 6H₂O. m(Al) = 0,5 g.
Keq= (products)/ (reactants)
Keq= ( [NO]^2 x [Cl2]) / ( [NOCl]^2)
Keq= ( (0.02)^2 x (0.01) ) / (0.5)^2= 1.6 x 10-5
The valence of lead is 4.
Hence the name of the compound is called Lead (IV) oxide.
<h3>Further explanation</h3>
Given
PbO₂ compound
Required
The valence of Pb
Solution
The oxidation number of element O in the compound = -2, except for OF₂ the oxidation state = + 2 and the peroxides (Na₂O₂, BaO₂) the oxidation state = -1 and superoxide, for example KO₂ = -1/2.
The oxidation state in the uncharged compound = 0,
So The oxidation state of Pb :
Pb + 2.(-2) = 0
Pb - 4 = 0
Pb = +4
Answer is: 25.84 milliliters of sodium metal.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.
d(Na) = 0.97 g/mL; density of sodim.
m(NaOH) = 43.6 g; mass of sodium hydroxide.
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 43.6 g ÷ 40 g/mol.
n(NaOH) =1.09 mol; amount of sodium hydroxide.
From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).
n(Na) = 1.09 mol.
m(Na) = 1.09 mol · 23 g/mol.
m(Na) = 25.07 g; mass of sodium.
V(Na) = m(Na) ÷ d(Na).
V(Na) = 25.07 g ÷ 0.97 g/mL.
V(Na) = 25.84 mL.
