P L E A S E help! explain and give answers on how to do moles for chemistry

1. Silver acetate → 166.87 g/m, 2.50 moles, 417.2 g, 1.50×10²⁴ particles

2. Glucose → 180 g/mol, 1.8 moles, 324 g, 1.08×10²⁴ particles

3. Lead sulfide → 239.26 g/m, 0.522 moles, 125 g, 3.14×10²³ particles

4. Iron (III) Chloride → 162.2 g/m, 0.390 moles, 63.3 g,  2.35×10²³ particles

5. Aluminum sulfate → 342.14 g/m, 1.56 mol, 533.7 g, 9.39×10²³ particles

6. Caffeine → 194 g/m, 7.17 moles, 1392 g, 4.32×10²⁴ particles

13.1, 83.9 L of N₂

13.2, 7.59 L of C₂H₆

13.3 232.8 L of SO₃

Explanation:

Molar mass Ag + 2 molar mass C + 3 molar mass H + 2 molar mass O ⇒ 107.87 g/m + 2 . 12 g/m + 3 . 1 g/m + 2 . 16 g/m =  166.87 g/m (molar mass)

If we have 2.50 moles , we have → 2.5 mol . 166.87 g/m = 417.2 g

1 mol of salt has 6.02×10²³ representative particles

2.5 moles of salt must have (2.5 . 6.02×10²³) / 1 = 1.50×10²⁴ particles

Molar mass → 180 g/mol

Let's convert the mass to moles → 324 g . 1 mol / 180 g = 1.8 moles

To determine number of particles → 1.8 mol . 6.02×10²³ particles / 1 mol =

1.08×10²⁴ particles

207.2 g/m + 32.06 g/m = 239.26 g/m

Mass to moles → 125 g . 1 mol / 239.26 g = 0.522 moles

0.522 moles . 6.02×10²³ particles / 1 mol = 3.14×10²³ representative particles.

Molar mass = Molar mass Fe + 3 Molar mass Cl

55.85 g/m + 3 . 35.45 g/m = 162.2 g/m

With the representative particles, we determine the moles.

2.35×10²³ particles . 1 mol / 6.02×10²³ particles = 0.390 moles

0.390 mol . 162.2 g / 1 mol = 63.3 g

Molar mass → 2 molar mass Al + 3 molar mass S + 12 molar mass O

2. 26.98 g/m + 3 . 32.06 g/m + 12 . 16 g/m = 342.14 g/m

We determine mass → 342.14 g /m . 1.56 mol = 533.7 g

1.56 moles . 6.02×10²³ particles / 1 mol = 9.39×10²³ particles

Molar mass → 8 . 12 g/m + 10 . 1 g/m + 4 . 14 g/m + 2. 16 g/m = 194 g/m

We determine the moles, by the representative particles:

4.32×10²⁴ particles . 1mol / 6.02×10²³ = 7.17 moles

We convert the moles to mass, 7.17 mol . 194g / 1 mol = 1392 g

Excercise 13:

1. P . V = n .  R . T

V = n . R . T / P → 3.75 mol . 0.082L.atm /mol.K . 273K / 1 atm = 83.9L

2. P .V = n . R . T

V =  n . R . T / P → 0.339 mol . 0.082L.atm /mol.K . 273K / 1 atm = 7.59L

3. We determine the moles of SO₃ → mass / molar mass

835 g / 80.06 g/m = 10.4 moles

V =  n . R . T / P → 10.4 mol . 0.082L.atm /mol.K . 273K / 1 atm = 232.8L

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