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Answer:
263.1 is exactly three half-life, so the remaining portion is (1/2 x 1/2 x 1/2) of the original sample. That's 1/8 which is 12.5%
Answer:
116.5 g of SO₂ are formed
Explanation:
The reaction is:
S₈(g) + 8O₂(g) → 8SO₂ (g)
Let's identify the moles of sulfur vapor, by the Ideal Gases Law
We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K
5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K
(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈
Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide
Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles
We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g
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