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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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Explanation:

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Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

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4 0
3 years ago
Calculate in gramm the mass of 0.1 mole of hydrochloric acid(h=1 ,cl=35.5)​
babunello [35]

Answer:

mass of HCl = 3.65 g

Explanation:

Data Given:

Moles of hydrochloric acid HCl = 0.1 mole

Mass in grams of hydrochloric acid HCl = ?

Solution:

Mole Formula

                  no. of moles = Mass in grams / molar mass

To find Mass in grams rearrange the above Formula

                Mass in grams = no. of moles x molar mass . . . . . . . (1)

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Put values in equation 1

              Mass in grams = 0.1 mole x 36.5 g/mol

              Mass in grams = 3.65 g

mass of HCl = 3.65 g

6 0
3 years ago
How many moles of water as a gas can be formed 2.45 L
Alex17521 [72]

Answer:

0.11mole

Explanation:

Let us assume that the condition is at standard temperature and pressure(STP);

 Given parameters:

        Volume of water  = 2.45L

   Unknown:

       Number of moles found in this volume of water  = ?

Solution;

 At STP;

                 Number of moles  = \frac{volume of gas}{22.4}

 Input the parameters and solve;

                  Number of moles of water  = \frac{2.45}{22.4}   = 0.11mole

The number of moles of water found is  0.11mole

4 0
3 years ago
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