Answer:
the specimen has 11460 years old
Explanation:
if the live sample has as initial amount of Yo C14, the dead sample will have 0.25Yo of C14.
the rate of decay of radiactive matter over time is
- Y(t) = Yo * e (( - t * Ln2 ) / T)
∴ Y(t) = 0.25Yo; T = 5730
⇒ 0.25Yo = Yo * e (( - t * Ln2 ) / 5730 )
⇒ 0.25 = e (( - t * Ln2 ) / 5730 )
⇒ Ln(0.25) = ( - t * Ln2 ) / 5730
⇒ - 7943.466 = - t * Ln2
⇒ 7943.466 / Ln2 = t
⇒ t = 11460 year
<span>2Al + 3Br2 -------------> 2AlBr3
</span>3 g Al = 0.11 mol Al.
<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>
<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>
<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>
</span>
Answer:
Explanation:
<u>1. Number of moles of gasoline</u>
a) Convert 60.0 liters to grams
- mass = 0.77kg/liter × 60.0 liter = 46.2 kg
- 46.2kg × 1,000g/kg = 46,200g
b) Convert 46,200 grams to moles
- molar mass of C₈H₁₈ = 114.2 g/mol
- number of moles = mass in grams / molar mass
- number of moles = 46,200g / (114.2 gmol) = 404.55 mol
<u>2. Number of moles of carbon dioxide, CO₂ produced</u>
a) Balanced chemical equation (given):
- C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)
b) mole ratio:
- 1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x
Solve for x:
- x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂
<u> 3. Convert the number of moles of carbon dioxide to volume</u>
Use the ideal gas equation:
- R = 0.08206 (mol . liter)/ (K . mol)
Substitute and compute:
- V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm
Round to two significant figures (because the density has two significant figures): 79,000 liters ← answer
