I really need these answers to balance the net ionic equation or make it into one.
Cu(NO3)2 (aq) + NaOH (aq) = CuOH (s) + Na(NO3)2 (aq)
Cu(NO3)2 (aq) + NaI (aq) = CuI (s) + Na(NO3)2 (aq)
Cu(NO3)2 (aq) + Na3PO4 (aq) = CuPO4 (s) + Na3(NO3)2 (aq)
Fe(NO3)3 (aq) + NaOH (aq) = FeOH (s) + Na(NO3)3 (aq)
I'm sorry Suey but your equations don't make sense. I can't balance equations that have the wrong formula.
Let's balance the net ionic equations step by step:
1. Cu(NO3)2 (aq) + 2NaOH (aq) = Cu(OH)2 (s) + 2NaNO3 (aq)
Balancing the Cu atoms:
Cu(NO3)2 (aq) + 2NaOH (aq) = Cu(OH)2 (s) + 2NaNO3 (aq)
Balancing the NO3^- ions:
Cu(NO3)2 (aq) + 2NaOH (aq) = Cu(OH)2 (s) + 2NaNO3 (aq)
Balancing the Na atoms:
Cu(NO3)2 (aq) + 2NaOH (aq) = Cu(OH)2 (s) + 2NaNO3 (aq)
2. Cu(NO3)2 (aq) + 2NaI (aq) = CuI2 (s) + 2NaNO3 (aq)
Balancing the Cu atoms:
Cu(NO3)2 (aq) + 2NaI (aq) = CuI2 (s) + 2NaNO3 (aq)
Balancing the NO3^- ions:
Cu(NO3)2 (aq) + 2NaI (aq) = CuI2 (s) + 2NaNO3 (aq)
Balancing the Na atoms:
Cu(NO3)2 (aq) + 2NaI (aq) = CuI2 (s) + 2NaNO3 (aq)
3. Cu(NO3)2 (aq) + 2Na3PO4 (aq) = Cu3(PO4)2 (s) + 6NaNO3 (aq)
Balancing the Cu atoms:
Cu(NO3)2 (aq) + 2Na3PO4 (aq) = Cu3(PO4)2 (s) + 6NaNO3 (aq)
Balancing the NO3^- ions:
Cu(NO3)2 (aq) + 2Na3PO4 (aq) = Cu3(PO4)2 (s) + 6NaNO3 (aq)
Balancing the Na atoms:
Cu(NO3)2 (aq) + 2Na3PO4 (aq) = Cu3(PO4)2 (s) + 6NaNO3 (aq)
4. Fe(NO3)3 (aq) + 3NaOH (aq) = Fe(OH)3 (s) + 3NaNO3 (aq)
Balancing the Fe atoms:
Fe(NO3)3 (aq) + 3NaOH (aq) = Fe(OH)3 (s) + 3NaNO3 (aq)
Balancing the NO3^- ions:
Fe(NO3)3 (aq) + 3NaOH (aq) = Fe(OH)3 (s) + 3NaNO3 (aq)
Balancing the Na atoms:
Fe(NO3)3 (aq) + 3NaOH (aq) = Fe(OH)3 (s) + 3NaNO3 (aq)
Please note that these are the balanced net ionic equations for the given reactions. The coefficients have been adjusted to ensure mass and charge conservation.
To balance or write the net ionic equations for the given reactions, follow these steps:
1. Write the balanced molecular equation.
In a molecular equation, the reactants and products are written as undissociated compounds. Balancing this equation means ensuring that the number of atoms of each element is the same on both sides of the equation.
Cu(NO3)2 (aq) + 2NaOH (aq) → Cu(OH)2 (s) + 2NaNO3 (aq)
Cu(NO3)2 (aq) + 2NaI (aq) → CuI2 (s) + 2NaNO3 (aq)
Cu(NO3)2 (aq) + 3Na3PO4 (aq) → Cu3(PO4)2 (s) + 6NaNO3 (aq)
Fe(NO3)3 (aq) + 3NaOH (aq) → Fe(OH)3 (s) + 3NaNO3 (aq)
2. Write the complete ionic equation.
In a complete ionic equation, all soluble ionic compounds are written as separate ions.
Cu2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2OH-(aq) → Cu(OH)2(s) + 2Na+(aq) + 2NO3-(aq)
Cu2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2I-(aq) → CuI2(s) + 2Na+(aq) + 2NO3-(aq)
Cu2+(aq) + 2NO3-(aq) + 9Na+(aq) + 3PO43-(aq) → Cu3(PO4)2(s) + 6Na+(aq) + 6NO3-(aq)
Fe3+(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq) → Fe(OH)3(s) + 3Na+(aq) + 3NO3-(aq)
3. Write the net ionic equation.
In the net ionic equation, the ions that remain unchanged on both sides of the equation are eliminated. Only those ions that undergo a chemical change are included.
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
Cu2+(aq) + 2I-(aq) → CuI2(s)
Cu2+(aq) + 3PO43-(aq) → Cu3(PO4)2(s)
Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)
These are the balanced net ionic equations for the given reactions.