Change this full equation to ionic equation.

${\mathrm{Br}}_2\left(\mathrm{aq}\right)+\mathrm{KI}\left(\mathrm{aq}\right)\to 2\mathrm{KBr}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right)$.

Write ionic equation for this precipitation reaction.

${\mathrm{CuSO}}_4\left(\mathrm{aq}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{Cu}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)$.

Write ionic equation for this precipitation reaction.

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+2\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{PbI}}_2\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$.

Calculate the concentration, in $\mathrm{mol} {\mathrm{dm}}^{-3}$, of the following solution: (${\mathrm{A}}_{\mathrm{r}}$ values: $\mathrm{H}=1.0, \mathrm{Na}=23.0, \mathrm{O}=16.0$).

A solution of sodium hydroxide, $\mathrm{NaOH}$, containing $2.0 \mathrm{g}$ of sodium hydroxide in $50 {\mathrm{cm}}^3$ of solution.

Calculate the concentration, in $\mathrm{mol} {\mathrm{dm}}^{-3}$, of the following solution: (${\mathrm{A}}_{\mathrm{r}}$ values: $\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0$).

A solution of ethanoic acid, ${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}$, containing $12.0 \mathrm{g}$ of ethanoic acid in $250 {\mathrm{cm}}^3$ of solution.

The equation for the reaction of strontium hydroxide with hydrochloric acid is shown below.

$\mathrm{Sr}{\left(\mathrm{OH}\right)}_2+2\mathrm{HCl}\to {\mathrm{SrCl}}_2+2{\mathrm{H}}_2\mathrm{O}$

$25.0 {\mathrm{cm}}^3$ of a solution of strontium hydroxide was exactly neutralised by $15.00 {\mathrm{cm}}^3$ of $0.100 \mathrm{mol} {\mathrm{dm}}^{-3}$ hydrochloric acid. Calculate the concentration, in $\mathrm{mol} {\mathrm{dm}}^{-3}$, of the strontium hydroxide solution.

$20.0 {\mathrm{cm}}^3$ of a $0.400 \mathrm{mol} {\mathrm{dm}}^{-3}$ solution of sodium hydroxide was exactly neutralised by $25.25 {\mathrm{cm}}^3$ of sulphuric acid. Calculate the concentration, in $\mathrm{mol} {\mathrm{dm}}^{-3}$, of the sulphuric acid. The equation for the reaction is: 

${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$.