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### Question 26:
# Explanation:
## Step1: Find molar mass of $ N_2S_2 $
Molar mass of $ N $: $ 14.01 \, g/mol $, molar mass of $ S $: $ 32.07 \, g/mol $.
Molar mass of $ N_2S_2 = 2\times14.01 + 2\times32.07 = 28.02 + 64.14 = 92.16 \, g/mol $.
## Step2: Calculate mass using $ m = n \times M $
Given $ n = 1.54 \, mol $, $ M = 92.16 \, g/mol $.
$ m = 1.54 \, mol \times 92.16 \, g/mol \approx 141.93 \, g $.
# Answer:
Approximately $ 142 \, g $ (or $ 141.93 \, g $)

### Question 27:
# Explanation:
## Step1: Find molar mass of $ NaOH $
Molar mass of $ Na $: $ 22.99 \, g/mol $, $ O $: $ 16.00 \, g/mol $, $ H $: $ 1.008 \, g/mol $.
Molar mass of $ NaOH = 22.99 + 16.00 + 1.008 = 39.998 \, g/mol \approx 40.00 \, g/mol $.
## Step2: Calculate moles using $ n = \frac{m}{M} $
Given $ m = 12.5 \, g $, $ M = 40.00 \, g/mol $.
$ n = \frac{12.5 \, g}{40.00 \, g/mol} = 0.3125 \, mol $.
# Answer:
$ 0.3125 \, mol $ (or $ 0.313 \, mol $)

### Question 28:
# Explanation:
1. **Count atoms on reactant side**:
   - $ CH_4 $: 1 C, 4 H; $ O_2 $: 2 O (from 1 $ O_2 $ molecule? Wait, no—wait the diagram: Left side: 1 $ CH_4 $ (1 C, 4 H) + 1 $ O_2 $ (2 O). Wait, no, let's re-examine the diagram. Wait, the reactants: $ CH_4 $ (1 C, 4 H) and $ O_2 $ (let's count the $ O_2 $ molecules. Wait the $ O_2 $ is a double bond, so 1 $ O_2 $ molecule (2 O). Products: $ CO_2 $ (1 C, 2 O) and $ H_2O $ (1 O, 2 H). Wait, no—wait the product side: 1 $ CO_2 $ (1 C, 2 O) and 1 $ H_2O $ (1 O, 2 H). Wait, that can't be. Wait, no—wait the diagram: Let's count atoms:

- Reactants:
     - $ CH_4 $: C=1, H=4.
     - $ O_2 $: Let's see the $ O_2 $ structure: the diagram shows $ O=O $ with two lone pairs on each O. So 1 $ O_2 $ molecule (O=2).

- Products:
     - $ CO_2 $: C=1, O=2.
     - $ H_2O $: O=1, H=2.

Wait, that's not balanced. Wait, no—maybe the number of $ O_2 $ and $ H_2O $ molecules is wrong. Wait, the correct reaction is $ CH_4 + 2 O_2
ightarrow CO_2 + 2 H_2O $. So in the diagram, the reactant $ O_2 $ should be 2 molecules (so 4 O), and product $ H_2O $ should be 2 molecules (so 2 O and 4 H). Wait, the diagram as drawn: reactants: 1 $ CH_4 $, 1 $ O_2 $; products: 1 $ CO_2 $, 1 $ H_2O $. That's unbalanced. So to balance, we need:

- Reactants: 1 $ CH_4 $, 2 $ O_2 $ (so 4 O).
   - Products: 1 $ CO_2 $, 2 $ H_2O $ (so 2 O from $ CO_2 $, 2 O from 2 $ H_2O $; 4 H from 2 $ H_2O $).

But the diagram shows 1 $ O_2 $ and 1 $ H_2O $. So we need to adjust the number of $ O_2 $ and $ H_2O $ molecules. Wait, the diagram: Let's look at the symbols. The $ O_2 $ is a single molecule (O=O), and $ H_2O $ is a single molecule (O with two H). So to balance:

- C: 1 (reactant) = 1 (product, $ CO_2 $) → balanced.
   - H: 4 (reactant, $ CH_4 $) → needs 4 H in products. Each $ H_2O $ has 2 H, so need 2 $ H_2O $ molecules (2×2=4 H).
   - O: Reactant $ O_2 $: let $ x $ be number of $ O_2 $ molecules. Product $ CO_2 $ has 2 O, 2 $ H_2O $ has 2×1=2 O. Total O: 2 + 2 = 4. So $ x \times 2 = 4 $ → $ x = 2 $.

So in the diagram, the reactant $ O_2 $ should be 2 molecules (so two $ O=O $ structures), and product $ H_2O $ should be 2 molecules (two $ H_2O $ structures). Wait, but the diagram as drawn has 1 $ O_2 $ and 1 $ H_2O $. So to balance, we need to add another $ O_2 $ molecule (so total 2 $ O_2 $) and another $ H_2O $ molecule (total 2 $ H_2O $).

Wait, the problem says "complete the particle model so that matter is conserved". So we need to draw the correct number of each molecule. Let's re-express:

- Reactants: 1 $ CH_4 $, 2 $ O_2 $ (so two $ O=O $ molecules).
   - Products: 1 $ CO_2 $, 2 $ H_2O $ (so two $ H_2O $ molecules).

So in the diagram, the $ O_2 $ should be 2 molecules (so add one more $ O_2 $ molecule to the reactant side), and the $ H_2O $ should be 2 molecules (add one more $ H_2O $ molecule to the product side).

Wait, the original diagram:

- Reactants: $ CH_4 $ (1) + $ O_2 $ (1) → Products: $ CO_2 $ (1) + $ H_2O $ (1).

To balance:

- C: 1 = 1 ✔️
   - H: 4 vs. 2 → need 2 $ H_2O $ (4 H)
   - O: 2 vs. 2 (from $ CO_2 $) + 2 (from 2 $ H_2O $) = 4 → need 2 $ O_2 $ (4 O)

So the particle model should have:

- Reactants: 1 $ CH_4 $, 2 $ O_2 $ molecules (so two $ O=O $ structures).
   - Products: 1 $ CO_2 $, 2 $ H_2O $ molecules (so two $ H_2O $ structures).

So in the diagram, we need to add one more $ O_2 $ molecule (so total 2 $ O_2 $) and one more $ H_2O $ molecule (total 2 $ H_2O $).

So the correct particle model:

- Reactant side: $ CH_4 $ (1) + $ O_2 $ (2) (two $ O=O $ molecules).
   - Product side: $ CO_2 $ (1) + $ H_2O $ (2) (two $ H_2O $ molecules).

So in the diagram, we need to draw another $ O_2 $ molecule (next to the first $ O_2 $) and another $ H_2O $ molecule (next to the first $ H_2O $).

# Answer:
To balance, add **1 more $ O_2 $ molecule** (so total 2 $ O_2 $ molecules) to the reactant side and **1 more $ H_2O $ molecule** (so total 2 $ H_2O $ molecules) to the product side. This gives:
- Reactants: $ CH_4 $ (1) + $ O_2 $ (2)
- Products: $ CO_2 $ (1) + $ H_2O $ (2)
(Visually, in the diagram, duplicate the $ O_2 $ structure once on the reactant side and duplicate the $ H_2O $ structure once on the product side.)

### Question 29a:
# Explanation:
## Step1: Predict products (double replacement: $ NaBr + H_3PO_4
ightarrow Na_3PO_4 + HBr $)
Double replacement: cations ($ Na^+ $, $ H^+ $) and anions ($ Br^- $, $ PO_4^{3-} $) swap.
So products: $ Na_3PO_4 $ (aq) and $ HBr $ (g or aq? $ HBr $ is a strong acid, so aq).

## Step2: Balance the equation
Unbalanced: $ NaBr + H_3PO_4
ightarrow Na_3PO_4 + HBr $
- $ Na $: 1 vs. 3 → put 3 in front of $ NaBr $.
- $ Br $: 3 (from 3 $ NaBr $) vs. 1 (from $ HBr $) → put 3 in front of $ HBr $.
- $ H $: 3 (from $ H_3PO_4 $) vs. 3 (from 3 $ HBr $) → balanced.
- $ PO_4^{3-} $: 1 vs. 1 → balanced.

Balanced equation: $ 3 NaBr_{(aq)} + H_3PO_{4(aq)}
ightarrow Na_3PO_{4(aq)} + 3 HBr_{(aq)} $

# Answer:
$ 3 NaBr_{(aq)} + H_3PO_{4(aq)}
ightarrow Na_3PO_{4(aq)} + 3 HBr_{(aq)} $

### Question 29b:
# Explanation:
## Step1: Predict products (double replacement: $ Pb(NO_3)_2 + KOH
ightarrow Pb(OH)_2 + KNO_3 $)
Cations: $ Pb^{2+} $, $ K^+ $; anions: $ NO_3^- $, $ OH^- $.
$ Pb(OH)_2 $ is insoluble (precipitate), $ KNO_3 $ is soluble (aq).

## Step2: Balance the equation
Unbalanced: $ Pb(NO_3)_2 + KOH
ightarrow Pb(OH)_2 + KNO_3 $
- $ Pb $: 1 vs. 1 → balanced.
- $ NO_3^- $: 2 vs. 1 → put 2 in front of $ KNO_3 $.
- $ K $: 1 (from $ KOH $) vs. 2 (from 2 $ KNO_3 $) → put 2 in front of $ KOH $.
- $ OH^- $: 2 (from 2 $ KOH $) vs. 2 (from $ Pb(OH)_2 $) → balanced.

Balanced equation: $ Pb(NO_3)_{2(aq)} + 2 KOH_{(aq)}
ightarrow Pb(OH)_{2(s)} + 2 KNO_{3(aq)} $

# Answer:
$ Pb(NO_3)_{2(aq)} + 2 KOH_{(aq)}
ightarrow Pb(OH)_{2(s)} + 2 KNO_{3(aq)} $

### Question 29c:
# Explanation:
## Step1: Predict products (double replacement: $ CaCl_2 + K_2CO_3
ightarrow CaCO_3 + KCl $)
Cations: $ Ca^{2+} $, $ K^+ $; anions: $ Cl^- $, $ CO_3^{2-} $.
$ CaCO_3 $ is insoluble (precipitate), $ KCl $ is soluble (aq).

## Step2: Balance the equation
Unbalanced: $ CaCl_2 + K_2CO_3
ightarrow CaCO_3 + KCl $
- $ Ca $: 1 vs. 1 → balanced.
- $ CO_3^{2-} $: 1 vs. 1 → balanced.
- $ K $: 2 (from $ K_2CO_3 $) vs. 1 (from $ KCl $) → put 2 in front of $ KCl $.
- $ Cl $: 2 (from $ CaCl_2 $) vs. 2 (from 2 $ KCl $) → balanced.

Balanced equation: $ CaCl_{2(aq)} + K_2CO_{3(aq)}
ightarrow CaCO_{3(s)} + 2 KCl_{(aq)} $

# Answer:
$ CaCl_{2(aq)} + K_2CO_{3(aq)}
ightarrow CaCO_{3(s)} + 2 KCl_{(aq)} $

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QUESTION IMAGE

Question

Explanation:

Question 26:

Step1: Find molar mass of \( N_2S_2 \)

Step2: Calculate mass using \( m = n \times M \)

Step1: Find molar mass of \( NaOH \)

Step2: Calculate moles using \( n = \frac{m}{M} \)

Answer:

Question 27: