The highest atom economy
2CO + O₂ ⇒ 2CO₂
<h3>Further explanation</h3>
Given
The reaction for the production of CO₂
Required
The highest atom economy
Solution
In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy
of the reaction
the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted
The general formula:
Atom economy = (mass of useful product : mass of all reactants/products) x 100
<em>or
</em>
Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100
So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C
Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Answer:
the chemical equation of ethanol as fuel is
C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)
the preparation process of ethanol from cassava is
cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol
Answer;
= 64561.95 g/mole
Explanation;
mass of Fe in 100g = .346g
= .346 / 55.8452 moles
= 0.0061957 moles
These represent 4 moles of Fe in the molecule so moles of hemaglobin
= 0.0061957/4
= 0.0015489 moles
these are in 100 g so mass of 1 mole = 100 / 0.0015489
= 64561.95 g / mole
molar mass of hemoglobin = 64561.95 g/mole
