Chemistry, 18.11.2019 01:31 nails4life324

What information does a net ionic equation give that a complete ionic equation does not show

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Answer from: Beth5225

The net ionic equation shows the chemical species involve in chemical reaction while the complete ionic equation also involve the spectator ions.

Explanation:

The net ionic equation shows the chemical species involve in chemical reaction while the complete ionic equation also involve the spectator ions. The net ionic equation state that whether precipitation is occur or not. If the chemical reaction have no net ionic equation it means all are in aqueous from not precipitation occur.

Consider the following example:

Balanced Chemical equation:

HClO₂(aq) + NaOH(aq) → H₂O(l) + NaClO₂ (aq)

Ionic equation:

H⁺(aq) + ClO₂⁻(aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + ClO₂⁻ (aq)

Net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The ClO₂⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

You know how the first one if filled up like the objects do the same but with the other categories

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