Answer:
For 1: The correct answer is False.
For 2: The correct answer is True.
For 3: The correct answer is True.
For 4: The correct answer is False.
For 5: The correct answer is True.
Step-by-step explanation:
Net ionic equation of any reaction does not include any spectator ions. If no net ionic equation is formed, it is said that no reaction has occurred.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.
- For 1: Lead(II) nitrate and sodium chloride
The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:
![Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/wb3kc8g1p32d9jntnu0uv2dl23lgs79ix8.png)
Ionic form of the above equation follows:
![Pb^(2+)(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/wmo9avn5wsguug3gsjj3dt9jghzhf4lk0t.png)
As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
![Pb^(2+)(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)](https://img.qammunity.org/2020/formulas/chemistry/college/vto43y8vtu1o57we2gnd4wyl90cid0v5wb.png)
Hence, the correct answer is False.
- For 2: Sodium bromide and hydrochloric acid
The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:
![NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/kdrnsr3942zrczyqz41slaq7h6mazcbt43.png)
Ionic form of the above equation follows:
![Na^(+)(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/4xdy83m0tw0xj6bhtesmw7zmun5alcib1z.png)
There are no spectator ions in the equation. So, the above reaction is the net ionic equation.
Hence, the correct answer is True.
- For 3: Nickel (II) chloride and lead(II) nitrate
The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:
![Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/kflpuctf8rtvopxjw2eiorgibrwo8xuoxn.png)
Ionic form of the above equation follows:
![Pb^(2+)(aq.)+2NO_3^-(aq.)+Ni^(2+)(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^(2+)(aq.)+2NO_3^-(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/ptmg6a2xnwbp8j5f7uykw1xf7hndiethhh.png)
As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
![Pb^(2+)(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)](https://img.qammunity.org/2020/formulas/chemistry/college/vto43y8vtu1o57we2gnd4wyl90cid0v5wb.png)
Hence, the correct answer is True.
- For 4: Magnesium chloride and sodium hydroxide
The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:
![MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/vz652qv4jk76ph9abar2c1qdigbug5hjr1.png)
Ionic form of the above equation follows:
![Mg^(2+)(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/7v2mxuy8vjpo000sp8y75f9i7foruezsrg.png)
As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
![Mg^(2+)(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)](https://img.qammunity.org/2020/formulas/chemistry/college/jygzazf5ej2z4kfyzltelz4qhco7tmr65s.png)
Hence, the correct answer is False.
- For 5: Ammonium sulfate and barium nitrate
The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:
![(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/dy36g45d6kc70s012b27xw1ho9mue604rv.png)
Ionic form of the above equation follows:
![2NH_4^(+)(aq.)+SO_4^(2-)(aq.)+Ba^(2+)(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)](https://img.qammunity.org/2020/formulas/chemistry/college/8ik2dqv6jaw7b25vlzfkq766oxcm3mrlbj.png)
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
![Ba^(2+)(aq.)+SO_4^(2-)(aq.)\rightarrow BaSO_4(s)](https://img.qammunity.org/2020/formulas/chemistry/college/nj83nintqvns6yjz3d9gajs8ejh9ym5t52.png)
Hence, the correct answer is True.