List the spectator ions in the following reaction.

ba+2(aq) + 2 oh-(aq) + 2 na+(aq) + co3-2(aq) → 2 na +(aq) + 2 oh-(aq) + baco3(s)

2Na⁺ (aq) and 2OH⁻(aq)

Explanation:

Spectator ions:

Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.

Given ionic equation:

Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)

In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq)  net ionic equation can be written as,

Net ionic equation:

Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml