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A 20 L sample of the gas contains 8.3 mol N₂.
According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant
<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁
<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁
___________
<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L
<em>n</em>₂ = ?; <em>V</em>₂ = 20 L
∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol
Answer: m = 24.31 g/mol · 1.13 mol
Explanation: 2 mol HCl use 1 mol Mg.
Magnesium is used 0.5 · 2.26 mol = 1.13 mol
M(Mg) = 24.31 g/ mol
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
Answer:
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