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Sagot :
To determine the net ionic equation from the given total ionic equation, follow these steps:
1. Write the total ionic equation:
[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]
2. Identify the spectator ions:
Spectator ions are ions that do not participate in the actual reaction and remain unchanged on both sides of the equation. In this equation, the spectator ions are:
- [tex]\( \text{NO}_3^- \)[/tex]: nitrate ion appears unchanged on both sides.
- [tex]\( \text{Na}^+ \)[/tex]: sodium ion appears unchanged on both sides.
3. Remove the spectator ions from the equation:
By canceling the common ions on both the reactant and product sides:
[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]
After removing 2 [tex]\( \text{NO}_3^- \)[/tex] and 2 [tex]\( \text{Na}^+ \)[/tex]:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
4. Write the net ionic equation:
The remaining ions form the net ionic equation:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
Thus, the correct net ionic equation is:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
So, the correct option for the net ionic equation is:
[tex]\( \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \)[/tex].
1. Write the total ionic equation:
[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]
2. Identify the spectator ions:
Spectator ions are ions that do not participate in the actual reaction and remain unchanged on both sides of the equation. In this equation, the spectator ions are:
- [tex]\( \text{NO}_3^- \)[/tex]: nitrate ion appears unchanged on both sides.
- [tex]\( \text{Na}^+ \)[/tex]: sodium ion appears unchanged on both sides.
3. Remove the spectator ions from the equation:
By canceling the common ions on both the reactant and product sides:
[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]
After removing 2 [tex]\( \text{NO}_3^- \)[/tex] and 2 [tex]\( \text{Na}^+ \)[/tex]:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
4. Write the net ionic equation:
The remaining ions form the net ionic equation:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
Thus, the correct net ionic equation is:
[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]
So, the correct option for the net ionic equation is:
[tex]\( \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \)[/tex].
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