You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of
carrying out all of the normal metabolic reactions) is incubated with 14C-labeled glucose at carbon 1. Based on your knowledge of glycolysis, where does the 14C-label end up in pyruvate?
1 answer:
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Answer:
(a) 0.779 mol Cr₂O₃
(b) 118 g Cr₂O₃
Explanation:
Step 1: Convert 156 g Cr₂S₃ to moles
Cr Molar Mass - 52.00 g/mol × 2 = 104.00 g/mol
S Molar Mass - 32.07 g/mol × 3 = 96.21 g/mol
156 g Cr₂S₃ ÷ 200.21 g/mol = 0.779182 mol Cr₂S₃
Step 2: Find conversion from Cr₂S₃ to Cr₂O₃
1 mol Cr₂S₃ equals 1 mol Cr₂O₃
Step 3: Use Dimensional Analysis
0.779182 mol Cr₂S₃ · = 0.779182 mol Cr₂O₃
0.779182 mol Cr₂O₃ ≈ 0.779 mol Cr₂O₃ (3 significant figures)
Step 4: Convert moles to grams
Cr Molar Mass - 52.00 g/mol × 2 = 104.00 g/mol
O Molar Mass - 16.00 g/mol × 3 = 48.00 g/mol
0.779182 mol Cr₂O₃ · 152.00 g/mol = 118.436 g Cr₂O₃
118.436 g Cr₂O₃ ≈ 118 g Cr₂O₃ (3 significant figures)