Iron-59 which is used to diagnose anaemia has a half life of 45days, what fraction of it is left in 90 days
Answer:
0.25 of the original
Explanation:
(0.5)^(90/45)
The fraction of iron-59 after 90 days is 0.25.
Iron-59 decays following first-order kinetics. Given the half-life ([tex]t_{1/2}[/tex]) of 45 days, we can calculate the rate constant (k) using the following expression.
[tex]k = \frac{ln2}{t_{1/2}} =\frac{ln2}{45day} = 0.015 d^{-1}[/tex]
For first-order kinetics, we can find the fraction of Fe ([Fe]/[Fe]₀) after and an elapsed time (t) of 90 days, using the following equation.
[tex]\frac{[Fe]}{[Fe]_0} =e^{-k \times t } = e^{-0.015 d^{-1} \times 90 d } \approx 0,25[/tex]
The fraction of iron-59 after 90 days is 0.25.
You can learn more about kinetics here:
https://brainly.com/question/2284525
The student is now told that the four solids, in no particular order, are calcium bromide (CaBr2), sugar (C6H12O6), benzoic acid (C6H5COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive. To rank items as equivalent, overlap them.
Answer:
Explanation:
Both calcium bromide and sodium bromide are ionic in nature so they will generate ions when dissolved in water .
CaBr₂ = Ca⁺² + 2Br⁻¹
NaBr = Na⁺ + Br⁻
Calcium bromide will generate greater number of ions per molecule according to chemical equation given above .
Benzoic acid ionizes but partially . Suger is a purely covalent compound so it does not generate ions at all .
In view of statement written above , the compounds can be arranged as follows according to conductivity from most to least conductive .
calcium bromide > sodium bromide > benzoic acid > sugar .