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

a

explanation:

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be