Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
Answer:
Since molarity is defined as moles of solute per liter of solution, we need to find the number of moles of nitric acid, and the volume of solution.
molar mass of nitric acid (HNO3) = 1 + 14 + (3x16) = 15 + 48 = 63 g/mole
1.50 g/ml x 1000 ml = 1500 g/liter
1500 g/liter x 0.90 = 1350 g/liter of pure HNO3 (the 0.9 is to correct for the fact that it is 90% pure)
1350 g/liter x 1 mole/63 g = 21.43 moles/liter = 21 Molar HNO3
= 21 Molar of HNO3
First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts
Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
