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Greeley [361]
3 years ago
12

Predict the products of each of these reactions and write balanced complete ionic and net ionic equations for each. If no reacti

on occurs, write NOREACTION .
Part A : LiCl(aq)+AgNO3(aq)→

Express your answer as a complete ionic equation. Identify all of the phases in your answer. Enter noreaction if no reaction occurs.

Part B: Express your answer as a net ionic equation. Identify all of the phases in your answer. Enter noreaction if no reaction occurs.

Part C: H2SO4(aq)+Li2SO3(aq)→

Express your answer as a complete ionic equation. Identify all of the phases in your answer. Enter noreaction if no reaction occurs.

Part D: Express your answer as a net ionic equation. Identify all of the phases in your answer. Enter noreaction if no reaction occurs.

Part E: HClO4(aq)+Ca(OH)2(aq)→

Express your answer as a molecular equation. Identify all of the phases in your answer. Enter noreaction if no reaction occurs.

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Express your answer as a complete ionic equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.

Part G:

Express your answer as a net ionic equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Express your answer as a complete ionic equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.

Part I: Express your answer as a net ionic equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.
Chemistry
1 answer:
Bumek [7]3 years ago
8 0

Answer:

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq)  →  AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq)  + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq)  →  AgCl(s) + Li⁻(aq)  + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) →  AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq)  →  Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq)  + SO²⁻₄(aq) + 2Li⁺(aq)  + SO₃²⁻(aq)  →  2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq)  →  SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq)  →  Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq)  →  2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq)  →   LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq)  →   3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq)  →   3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) +  3OH⁻ (aq)  →    Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq)  → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq)   → Hg₂Cl₂ (s)

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<h3>What is a Percent yield</h3>

A percent yield of a substance measures the amount of the substance actually obtained as a percentage ratio of expected yield.

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<h3>How to calculate the mass of aluminium obtained from bauxite </h3>

From the data given:

40 % of the bauxite is converted to aluminium oxide.

Volume of bauxite = 1 m^3

40 % of 1 m^3 = 0.4 m^3

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mass of aluminium oxide = 0.4 × 3965 kg

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molar mass of aluminium oxide = 102 g

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Percentage mass of aluminium in aluminium oxide = 54/102 × 100

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which atoms attains a stable valence electron configuration by bonding with another atom 1) neon 2) radon 3) helium 4) hyrdogen
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Well, when an atom attains a stable valence electron, it means that the outer electrons are complete and so cannot attain any more electrons. For the first shell, it is complete when it has 2 electrons, the second shell is complete when it has 8 electrons, all the other shells also have a particular number when complete. Anyway, i believe the answer is HYDROGEN because when HYDROGEN combines with another atom of HYDROGEN, the outer shell is completed. This is because HYDROGEN has only 1 electron. If the two HYDROGENS, which both have 1 electron combine, they make the electrons 2, which is complete for the first shell, HYDROGEN ends in the first shell. Since the electrons become 2, the shell is at stable valence. In all the other options, this happens;
NEON- It has 10 electrons, 2 in the first shell and 8 in the second. So the the shells are already complete, so it can't bond with any thing, which is completely against the question.
RADON- Radon has 86 electrons.
HELIUM- Helium has 2 electrons, so the shell is already full, and cannot bond, so it goes against the question. The question says BY BONDING.
 So the answer is definitely 4) HYDROGEN
Hope i helped. Have a nice day, by the way, i'm very sure it's hydrogen.

3 0
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