Answer:
2.99×10²⁵ molecules of CO₂ are produced
Explanation:
Decomposition reaction is:
Ca(HCO₃)₂ => CaO(s) + 2CO₂(g) + H₂O(g)
Ratio is 1:2. Let's make a rule of three:
1 mol of bicarbonate can produce 2 moles of CO₂
Therefore, 24.9 moles of bicarbonate may produce, 49.8 moles (24.9 .2 )/1
Let's determine the number of molecules
1 mol has 6.02×10²³ molecules
49.8 moles must have (49.8 . 6.02×10²³) / 1 = 2.99×10²⁵ molecules
The correct answer is A I believe
Answer:
32.4 mol
Explanation:
Given data:
Number of moles of C atom present = ?
Number of moles of glucose = 5.4 mol
Solution:
Glucose formula = C₆H₁₂O₆
There are 6 moles of C atoms are present in one mole of glucose.
In 5.4 moles of glucose:
5.4 mol × 6 = 32.4 mol
Answer:
1/3
Explanation:
Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.
Pyruvate is converted to acetyl-CoA in the reaction given below:
Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂
1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.
Also,
2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).
Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.
Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)
Thus,
<u>Fraction = 2/6 = 1/3</u>