the primary consumer in those photos would be C
Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%
Yes it is available. It will continue catalyzing the reactions until it becomes completely consumed. That's how enzymes work. They work and are eventually consumed in the process completely without altering the reaction in any way other than speeding it up.
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An oxide is a molecule that has 2 Oxygen atoms in its empirical formula. Stoichiometry would be used as the ratio of Oxygen to Dioxides is 3:2. So 52.29 would be multiplied by 3/2. The answer is 78.435 Mol of CO2 are formed.
Answer:
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