Question

CuCl2 + KOH--> Розвязати,та записати йонне,та скорочене йонне рывняння. AgNO3+BaCl2--> Na2CO3+HNO3-->

Answer 1

CuCl2 + 2KOH = Cu(OH)2 (осадок) + 2KCl
Cu + 2Cl + 2K + 2OH = Cu(OH)2 + 2K + 2Cl
Cu + 2OH = Cu(OH)2

2AgNO3 + BaCl2 = 2AgCl (осадок) + Ba(NO3)2
2Ag + 2NO3 + Ba + 2Cl = 2AgCl (осадок) + Ba + 2NO3
2Ag + 2Cl = 2AgCl (осадок)
Ag + Cl = AgCl (осадок)

Na2CO3 + 2HNO3 = 2NaNO3 + H2CO3
2Na + CO3 + 2H + 2NO3 = 2Na + 2NO3 = 2H + CO3
сокращается все, не остается ничего

Answer 2

CuCl2 + KOH

The reaction between CuCl2 (copper(II) chloride) and KOH (potassium hydroxide) can be written as follows:

CuCl2 + 2KOH -> Cu(OH)2 + 2KCl

In this reaction, copper(II) chloride reacts with potassium hydroxide to form copper(II) hydroxide and potassium chloride. The balanced ionic equation for this reaction is:

Cu^2+ + 2OH^- + 2K^+ + 2Cl^- -> Cu(OH)2 + 2K^+ + 2Cl^-

And the simplified ionic equation is:

Cu^2+ + 2OH^- -> Cu(OH)2

AgNO3 + BaCl2

The reaction between AgNO3 (silver nitrate) and BaCl2 (barium chloride) can be written as follows:

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

In this reaction, silver nitrate reacts with barium chloride to form silver chloride and barium nitrate. The balanced ionic equation for this reaction is:

Ag^+ + NO3^- + Ba^2+ + 2Cl^- -> AgCl + Ba^2+ + 2NO3^-

And the simplified ionic equation is:

Ag^+ + Cl^- -> AgCl

Na2CO3 + HNO3

The reaction between Na2CO3 (sodium carbonate) and HNO3 (nitric acid) can be written as follows:

Na2CO3 + 2HNO3 -> 2NaNO3 + H2O + CO2

In this reaction, sodium carbonate reacts with nitric acid to form sodium nitrate, water, and carbon dioxide. The balanced ionic equation for this reaction is:

2Na^+ + CO3^2- + 2H^+ + 2NO3^- -> 2Na^+ + 2NO3^- + H2O + CO2

And the simplified ionic equation is:

CO3^2- + 2H^+ -> H2O + CO2

I hope this helps! Let me know if you have any further questions.