The answer is: the mass of oxygen is 16.95 grams.
The overall balanced photosynthesis reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
m(C₆H₁₂O₆) = 15.90 g; mass of glucose.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.
n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.
From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.
n(O₂) = 6 · 0.088 mol.
n(O₂) = 0.53 mol; amount of oxygen.
m(O₂) = 0.53 mol · 32.00 g/mol.
m(O₂) = 16.95 g; mass of oxygen.
Answer:
The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.
Explanation:
The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.
Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i
Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.
It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.
the number of moles of oxygen required are 0.08 mol. The volume of oxygen that is required to react can be calculated by the formula shown below. Substitute the values in equation (II). Hence, the volume of oxygen required to react with 3.6 L hydrogen is 1.8L . I hope this helps if not I’m sorry
Answer:
Animal cells have centrosomes (or a pair of centrioles), and lysosomes, whereas plant cells do not. Plant cells have a cell wall, chloroplasts, plasmodesmata, and plastids used for storage, and a large central vacuole, whereas animal cells do not.
The percent yield of CO₂ is 93.3%.
<h3>What is the percent yield of CO₂?</h3>
The percent yield of a substance is given as follows:
- Percent yield = actual yield/theoretical yield * 100 %
The equation of the reaction is used to determine the theoretical yield.
- NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
Moe ratio of sodium bicarbonate and CO₂ is 1 : 1.
Given that sodium bicarbonate is the limiting reactant, the theoretical yield of CO₂ will be:
Moles of NaHCO₃ reacting = 2.01/84 = 0.0239 moles
Theoretical yield of CO₂ = 0.0239 moles * 22.4L/mol = 0.536 L
Actual yield = 0.50 L
Percent yield = 0.50/0.536 * 100%
Percent yield = 93.3%
In conclusion, the percent yield is the ratio of the actual yield and theoretical yield.
<em>Note that the complete question is given below:</em>
<em>Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.50 L of at s.t.p.</em>
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