Sallyjoe9630

29.08.2021 •

Chemistry

What is the dimension analysis for 154.101 –23.22 =

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Ответ:

eme05

19.07.2021

Chemistry

Answer : The net ionic equation will be,

(a) $C_2O_4^{2-}(aq)+Ba^{2+}(aq)\rightarrow BaC_2O_2(s)$

(b) $Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$K_2C_2O_4(aq)+Ba(OH)_2(aq)\rightarrow 2KOH(aq)+BaC_2O_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+C_2O_4^{2-}(aq)+Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow BaC_2O_2(s)+2K^+(aq)+2OH^{-}(aq)$

In this equation, $K^+\text{ and }OH^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$C_2O_4^{2-}(aq)+Ba^{2+}(aq)\rightarrow BaC_2O_2(s)$

(b) The given balanced ionic equation will be,

$Pb(NO_3)_2(aq)+H_2SO_4(aq)\rightarrow 2HNO_3(aq)+PbSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2H^+(aq)+2NO_3^{-}(aq)$

In this equation, $H^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

$CaCO_3(s)+H_2SO_4(aq)\rightarrow CO_2(g)+CaSO_4(aq)+H_2O(l)$

The ionic equation in separated aqueous solution will be,

$CaCO_3(s)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow CO_2(g)+Ca^{2+}(aq)+SO_4^{2-}(aq)+H_2O(l)$

In this equation, $SO_4^{2-}$ is the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.The net ionic equation will be,

$CaCO_3(s)+2H^+(aq)\rightarrow CO_2(g)+Ca^{2+}(aq)+H_2O(l)$

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