We represent the species in the sodium iodide solution as Na+(aq) and I–(aq), and in the lead(II) nitrate solution as the species Pb2+(aq) and NO3–(aq). In the process of dissolving, the ions from the ionic solutes (sometimes referred to as salts) are split apart and hydrated (surrounded) by water molecules. Many ionic compounds are water soluble, so, for example, we can make solutions of sodium iodide, NaI, and lead(II) nitrate, Pb(NO3)2, by simply dissolving the ionic solids in water. (Note I left the mass in units of micrograms but could have also worked in grams but it would have been messier.) To check our answer note that the time calculated is less than a half-life and the mass decayed is less than half the mass originally present. Taking natural log to both sides of the equation: Now lets calculate the value for k and we will finally can compute the age of the sample: To find the mass decayed multiply by the atomic weight:Ġ.3544 x 10⁻³ in μmol x 238 g/ m in μmol = 8.434 x 10⁻² in μg ²³⁸ U decayed This means 0.3544 x 10⁻³ in μmol of ²³⁸ U decayed The hints given in the problem helps us greatly to determine N and N₀Īll the ²⁰⁶ Pb comes from ²³⁸ U and we know its mass = 0.730 in μg, so lets find moles ²³⁸ UĠ.730 ug x 1 mol ²⁰⁶ Pb/206 umol = 0.3544 x 10⁻³ umol Which is derived from Nt/N₀ = e^-kt for the half-life where Nt/N₀ = 0.5 We notice we dont have k, but this value can be obtained from t₁/₂ through N₀ = amount of ²³⁸ U originally present in μg Nt = amount of ²³⁸ U present at time t in μg ( μ =microgram) Nt/N₀ = e^-kt where k is the decay constant per year Here we will be using the radiactive decay equations : What is the relationship between the number of 206Pb atoms present and the amount of 238U that has decayed? How much 238U has not yet decayed? Did you find the ratio, Nt/N0, the amount of 238U remaining relative to the amount of 238U initially present? 238U decays to 206Pb through a series of 14 steps with an apparent half-life of 4.47×109 years. Approximately how old is the sample?Collect and OrganizeAssuming that the only loss of 238U is via radioactive decay and that all of the nuclides produced by the decay processes remain in the sample, the U-Pb radiometric dating method can be used to calculate the age of the zircon sample. They find 0.730 microgram of 206Pb for 1.000 microgram of 238U present. The U-Pb dating method relies on two separate decay chains, one of which is the uranium series from 238U to 206Pb, with a half-life of 4.47 billion years.Geologists unearth a sample of zircon that appears to be a closed system. Uranium-lead (U-Pb) dating of geological samples is one of the oldest and most refined radiometric dating methods, able to determine ages of about 1 million years to over 4.5 billion years with precision in the 0.1–1% range. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be: Now, we verify the amount of the elements, which must be equal on both sides. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is: In the given reaction:Īs we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The number of electrons must be the same, so the second equation must be multiplied by 3:īrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)įind more information about Balanced chemical equation here:Īt a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). So, it's going from +5 to -1, and the half-reactions are: The oxidation number of bromine in the reactant, knowing that the oxidation number of O is -2, is: When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:īrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)Īs we can see, the antimony is being oxidized (go from +3 to +5), and the Bromine is being reduced. BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)Īt a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons).īrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)
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