Chemistry, 09.12.2019 17:31 jbehrens6538

Write the net ionic equation to show the formation of a solid (insoluble salt) when the following solutions are mixed. write noreaction if there is no precipitate. express your answer as a chemical equation. identify all of the phases in your answer. enter noreaction if no precipitate is formed.1. ca(no3)2(aq) + na2so4(aq)2. kcl(aq) + pb(no3)2(aq)3. cacl2(aq) + (nh4)3(po4)(aq)4. na2so4(aq) + bacl2(aq)

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Answer from: nkidane2802

The net ionic equations are written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

For 1:

The chemical equation for the reaction of calcium nitrate and sodium sulfate is given as:

$Ca(NO_3)_2(aq.)+Na_2SO_4(aq.)\rightarrow CaSO_4(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ca^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$

For 2:

The chemical equation for the reaction of lead (II) nitrate and potassium chloride is given as:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)$

For 3:

The chemical equation for the reaction of calcium chloride and sodium phosphate is given as:

$3CaCl_2(aq.)+2(NH_4)_3PO_4(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4Cl(aq.)$

Ionic form of the above equation follows:

$3Ca^{2+}(aq.)+6Cl^-(aq.)+6NH_4^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4^+(aq.)+6Cl^-(aq.)$

As, ammonium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$3Ca^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)$

For 4:

The chemical equation for the reaction of barium chloride and sodium sulfate is given as:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the net ionic equations are written above.

Answer from: Quest

a plasma is a warm ionized gasoline consisting of approximately equal numbers of definitely charged ions and negatively charged electrons. the characteristics of plasmas are notably unique from the ones of ordinary neutral gases in order that plasmas are considered a awesome "fourth kingdom of remember."

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.