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Basile [38]
3 years ago
15

Enter a net ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II

) sulfate and aqueous potassium nitrate. Express your answer as a chemical equation. Identify all of the phases in your answer.
Chemistry
1 answer:
iogann1982 [59]3 years ago
7 0

<u>Answer:</u> The net ionic equation is written above

<u>Explanation:</u>

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.

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

Pb(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+PbSO_4(s)

Ionic form of the above equation follows:

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

Hence, the net ionic equation is written above.

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Answer:

ΔE = -2661 KJ/mole

ΔH = -2658 KJ/mole

Explanation:

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<u>First, to find ΔE:</u>

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2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction <u><em>PRODUCES</em></u> heat, which means ΔE is negative.

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<u>Second, to find ΔH:</u>

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Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.

ΔH = 2658 KJ(q) - (0)ΔV

ΔH = 2658 KJ - 0

ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.

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I hope this helped! Have a nice week.

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