Asked by chemdummy
what is the 1. chemical equation 2. complete ionic equation 3. net ionic equation include phase of HCH3COO+NaHCO3 pls correct my answers
1. CH3COOH(aq)+NaHCO3(aq)+CH3COONa(aq)+H2O(l)+CO2(g)
2. CH3COOH(aq)+OH-(aq)=CH3COO-(aq)+H2O(l)
3. OH(aq)+CO2(g)
1. CH3COOH(aq)+NaHCO3(aq)+CH3COONa(aq)+H2O(l)+CO2(g)
2. CH3COOH(aq)+OH-(aq)=CH3COO-(aq)+H2O(l)
3. OH(aq)+CO2(g)
Answers
Answered by
DrBob222
The complete molecular equation is correct if you take away the + sign and replace it with an arrow where the arrow should be.
2. The complete ionic equation is not right. It should be this way.
CH3COOH(aq) + Na^+(aq) + HCO3^-(aq) ==> CH3COO^-(aq) + Na^+(aq) + H2O()l) + CO3(g). All I've done is taken your molecular equation #1 and split each reactant/product into ions if they ionize.
3. Cancel the spectator ions (those that appear on both sides of the equation) and you're left with the net ionic equation.
CH3COOH(aq) + HCOL3^-(aq) ==> H2O(l) _ CO2(g) + CH3COO^-(aq)
2. The complete ionic equation is not right. It should be this way.
CH3COOH(aq) + Na^+(aq) + HCO3^-(aq) ==> CH3COO^-(aq) + Na^+(aq) + H2O()l) + CO3(g). All I've done is taken your molecular equation #1 and split each reactant/product into ions if they ionize.
3. Cancel the spectator ions (those that appear on both sides of the equation) and you're left with the net ionic equation.
CH3COOH(aq) + HCOL3^-(aq) ==> H2O(l) _ CO2(g) + CH3COO^-(aq)
Answered by
Nkosi mncedisi Sonnyboy
1. H2SO4 + 2NaHCO3= NaSO4 + 2CO2 + 2H2O
2. 2Na^+ + CO3^2- + 2H^+ SO4^2-= 2Na^+ + SO4^2- + 2H2O + 2CO2
3.HCO3^- + H^+ = H2O + CO2
2. 2Na^+ + CO3^2- + 2H^+ SO4^2-= 2Na^+ + SO4^2- + 2H2O + 2CO2
3.HCO3^- + H^+ = H2O + CO2
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