The answer is b. radon-222. The alpha decay means that it will emit an alpha particle when decays. The alpha particle has two protons and two neutrons. So Radium(88) minus two protons will become Radon(86). And the atomic mass will become 226-4=222.
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
The atoms in the first period have electrons in 1 energy level. The atoms in the second period have electrons in 2 energy levels. The atoms in the third period have electrons in 3 energy levels. The atoms in the fourth period have electrons in 4 energy levels.
To determine the relative atomic mass of thallium, we multiply the molar mass of the isotopes to their corresponding relative abundance. The molecular percentages should sum up to 1. In this case, we multiply 203 by 0.295 and 205 by 0.705 and add the answers of the two. The final atomic mass is 204.41 g/mol.
balanced equation =
3Cu(OH)2 + 2H3PO4 → Cu3(PO4)2 + 6H2O
