Is the equation for reaction 1 balanced? Explain your reasoning. If the reaction is not balanced, then state how you would balan
ce it. Then, provide the balanced equation.
1 answer:
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Answer:
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:
- Silver acetate → AgCH₃COO
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
- Glucose → C₆H₁₂O₆
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
- PbS → Lead sulfide → Molar mass = Molar mass Pb + Molar mass S
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.
- FeCl₃ → Iron(III) chloride
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
- Aluminum sulfate → Al₂(SO₄)₃
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
- Caffeine → C₈H₁₀N₄O₂
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
<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
- <u>For
isotope:</u>
Mass of isotope = 63.929 amu
Percentage abundance of isotope = 48.63 %
Fractional abundance of isotope = 0.4863
- <u>For
isotope:</u>
Mass of isotope = 65.926 amu
Percentage abundance of isotope = 27.90 %
Fractional abundance of isotope = 0.2790
- <u>For
isotope:</u>
Mass of isotope = 66.927 amu
Percentage abundance of isotope = 4.10 %
Fractional abundance of isotope = 0.0410
- <u>For
isotope:</u>
Mass of isotope = 67.925 amu
Percentage abundance of isotope = 18.75 %
Fractional abundance of isotope = 0.1875
- <u>For
isotope:</u>
Mass of isotope = 69.925 amu
Percentage abundance of isotope = 0.62 %
Fractional abundance of isotope = 0.0062
Putting values in equation 1, we get:
Hence, the average atomic mass of element Zinc is 65.40 amu.