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Greeley [361]
2 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]2 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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Suppose you have 81.8 g of Cd. How many Cd atoms are present? Report your answer in scientific notation using the format of 6.02
Lostsunrise [7]

Answer:

1. number of Cd atoms = 4.384 × 10²³

2. number of moles of Ti = 0.817 mol

3. molar mass of C₇H₆O₃ = 138.077 g/mol

4. moles of Sildenafil = 0.249 mmol

Explanation:

1. given data

mass of Cd = 81.8 g

number of Cd atoms = ?

molar mass of Cd = 112.411 g/mol

Solution

1st we find out the number of moles of Cd

<em>         number of moles = mass / molar mass </em>

 number of moles of Cd = 81.8 g / 112.411 g.mol⁻

 number of moles of Cd = 0.728 mol

Now we find out the number of Cd atoms

<em>       number of atoms = moles × 6.022 × 10²³</em>

 number of Cd atoms = 0.728 (6.022 × 10²³)

 number of Cd atoms = 4.384 × 10²³

 

2. Given data

Mass of Ti = 39.1 g

moles = ?

molar mass Ti = 47.867 g/mol

Solution

Now we will find out the number of moles

<em>         number of moles = mass / molar mass </em>

 number of moles of Ti = 39.1 g / 47.867 g.mol⁻

 number of moles of Ti = 0.817 mol

3. Given data

chemical formula of Salicylic acid  = C₇H₆O₃

molar mass of salicylic acid = ?

molar mass of carbon = 12.011

molar mass of hydrogen = 1.008

molar mass of oxygen = 15.999

<em>         molar mass of  molecule = number of moles of atoms × molar mass of atoms</em>

  molar mass of C₇H₆O₃ = 7 (12.011) + 6 (1) + 3 (16)

  molar mass of C₇H₆O₃ = 84.077 g + 6 g + 48 g

  molar mass of C₇H₆O₃ = 138.077 g/mol

4. Given data

molar mass of Sildenafil = 474.5828 g/mol

mass given = 0.118 g

moles of Sildenafil = ?

Solution

Now we find out the number of moles

<em>          number of moles = mass / molar mass </em>

<em>   </em>number of moles Sildenafil = 0.118 g / 474.5828 g/mol

  number of moles Sildenafil = 0.0003 mol

as this is very small amount we convert it into millimoles

           <em>millimoles = moles × 1000</em>

    millimoles of Sildenafil = 0.000249 × 1000

    millimoles of Sildenafil = 0.249 mmol

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