An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.
The stronger the attraction between elements, the shorter the bond length becomes; conversely, a weaker attraction results in a longer bond length. This attraction arises from differences in their electronegativities, which is the capacity of an element to draw electrons toward itself. According to periodic trends, electronegativity rises as you move left to right and bottom to top on the periodic table. Therefore, the order from the most electronegative to the least is: Cl > Br > I. As a result, the sequence by bond length from shortest to longest is: C-Cl > C-Br > C-I.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.
Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
7.35 moles of oxygen. Initially, for each mole of H₂CO₃, there are 3 moles of oxygen, as derived from the acid's formula. For 2.45 moles of the compound stated in the problem, which is carbonic acid, we calculate: If 1 mole of H₂CO₃ corresponds to 3 moles of oxygen, then for 2.45 moles of H₂CO₃, we have X moles of oxygen. Thus, X = (3 × 2.45) / 1 = 7.35 moles of oxygen.
What is being removed during the wash is the solvent.
