Answer:
Explanation:
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In this case, since the molar mass of the empirical formula is:
Thus, the ratio of the molecular to the empirical formula is:
Thus, the molecular formula is six times the empirical formula:
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Different radioactive nuclides decay into their respective daughter nuclides at distinct rates. Some of the nuclides decay briskly, while others decay gradually. The time it consumes for half of the parent nuclide in a radioactive sample to decay to the daughter nuclides is known as the half-life of the radioactive sample.
The nuclides, which decay briskly exhibit short half-lives and are very active. The half-life can be utilized to find the rates of radioactive decay. In the given question, the half-lives of various nuclides are given. So, the order to the most active (shortest half-life or largest number of decays per second) to least reactive (largest half-life or the smallest number of decays per second) is:
Tc-99m > Y-90 > In-111 > I-131
The answer is 1. Radiation
Answer:
2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20
Explanation:
Since there's no "?" shown in the equation, let's balance it and solve it entirely.
Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20
1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.
To do this, let's start by looking at the elements that are only presnet once on each side:
On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H → to balance it, you need 4 water molecules on the products side.
N₂ + 3 H₂ → 2 NH₃
number of moles of N₂ = 38 g / 28 g/mol = 1.357 mol
1 mole of N₂ gives 2 moles of NH₃
1.357 mol of N₂ will give ?? mole NH₃
= 2 x 1.357 = 2.7 mol NH₃
