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marishachu [46]
3 years ago
13

Write the net ionic equation for ammonium carbonate and lead(II) nitrate are combined. (Use the solubility rules provided in the

OWL Preparation Page to determine the solubility of compounds.)
Chemistry
1 answer:
sashaice [31]3 years ago
8 0

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

<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 ammonium carbonate and lead (II) nitrate is given as:

(NH_4)_2CO_3(aq.)+Pb(NO_3)_2(aq.)\rightarrow 2NH_4NO_3(aq.)+PbCO_3(s)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+CO_3^{2-}(aq.)+Pb^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow PbCO_3(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium 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.)+CO_3^{2-}(aq.)\rightarrow PbCO_3(s)

Hence, the net ionic equation is written above.

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The change from figure one to figure two was most likely caused by erosion. Erosion is the process of something being eroded by wind, water, or other natural agents.
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3 years ago
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Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and
tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of PCl_5 is 1.285 M.

<u>Explanation:</u>

The chemical equation for the decomposition of phosphorus pentachloride follows:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}

We are given:

K_c=0.042

[PCl_3]=0.18M

[Cl_2]=0.30M

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

0.042=\frac{0.18\times 0.30}{[PCl_5]}

[PCl_5]=1.285

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How many grams are in 3.78 x 10^22 molecules of SO2?
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Given the following information:Br2 bond energy = 193 kJ/mol F2 bond energy = 154 kJ/mol 1/2Br2(g) + 3/2F2(g) -&gt; BrF3(g) = –3
arlik [135]

Answer:

C) 712 KJ/mol

Explanation:

  • ΔH°r = Σ Eb broken - Σ Eb formed
  • 1/2Br2(g) + 3/2F2(g) → BrF3(g)

∴ ΔH°r = - 384 KJ/mol

∴ Br2 Eb = 193 KJ/mol

∴ F2 Eb = 154 KJ/mol

⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)

⇒ Σ Eb broken =  (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol

∴ Eb formed: Br-F

⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r

⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )

⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol

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Convert 2.50 x 10^5 seconds into years
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...........................

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