Question

Question #cb7dd

Answer 1

Alright, it's been a while since I've done net ionic equations, so pls call me out if I get something wrong.

1. `SO_4^-2+Ca^(+2)->CaSO_4(s)`
2. `Cu^(+2)+CO_3^-2->CuCO_3(s)`
3. `OH^(-)+H^(+)->H_2O(l)`
4. First, of all that equation is balanced incorrectly. It should be `Na_2CO_3+2HCl=2NaCl+H_2CO_3`. Secondly, since both products are soluble, there is no reaction, and hence no net ionic equation.
5. Firstly, MgOH is not a compound. Magnesium Hydroxide is `Mg(OH)_2`. Also, `KSO_4` is also not a compound. Potassium Sulfate is `K_2SO_4`. Using these correct compounds, the equation is balanced to read `2KOH+MgSO_4->Mg(OH)_2+K_2SO_4`. Using this correct equation, the net ionic equation is `2OH^(-)+Mg^(+2)->Mg(OH)_2(s)`
6. Firstly, `NaSO_4` is not a compound. Sodium Sulfate is `Na_2SO_4`. Really, there is no reaction occurring here at all, since the problem writer just switched the order of the reactants for the products. As such, there is no reaction and no net ionic equation
7. Just like the last one, the problem writer has just switched the order of the reactants for the products. As such, there is no reaction and no net ionic equation

Answer 2

"4. The greater part of `H_2CO_3` will disintegrate according to:
`H_2CO_3->H_2O+CO_2(g)`