The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
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Answer: It will take 11.775 seconds.
Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is
A = 4.π.r²
r for an alveolus would be: r = 0.00005m or r = 5.
m
Finding the area:
A = 4.3.14.(5.
)²
A = 3.14.
m²
The concentration change is to be 90% of the final, so
c = 0.9.3.14.
c = 28.26.
The oxygen diffusivity is 2.4.
m²/s, that means in 1 second 2.4.
of oxygen spread in one alveolus area. So:
1 second = 2.4.
m²
t seconds = 28.26.
m²
t = 
t = 11.775s
For a concentration change at the center to be 90%, it will take 11.775s.
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Answer:
7.5 mol
Step-by-step explanation:
2Al + 3Br₂ ⟶ 2AlBr₃
You want to convert moles of AlBr₃ to moles of Br₂.
The molar ratio is 3 mol Br₂:2 mol AlBr₃.
Moles of Br₂ = 5 × 3/2
Moles of Br₂ = 7.5 mol Br₂
You need 7.5 mol of Br₂ to produce 5 mol of AlBr₃.
Answer:
What is true of a solution of 1.0 M HCl(aq)? It contains 1.0 gram of HCl per 100 grams of water.
Explanation: