D = M/V
D = 0.10
V = 1000 cm^3
0.10 = M / 1000
M = 100
Porque la densidad es 10%, se puede usar 0.10 en este formulario para calcular el peso. No sé lo que es la unidad para el peso pero es 100.
Answer:
(molecular) 3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
(ionic) 3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
(net ionic) 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)
Explanation:
The molecular equation includes al the species in the molecular form.
3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
The ionic equation includes all the ions (species that dissociate in water) and the species that do not dissociate in water.
3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water. In does not include <em>spectator ions</em>.
3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
Answer:
the answer is 70.906 g/mol
Explanation: