Answer:
C. Particle size
Explanation:
The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not
Answer:
Electron transport is the process by NADH + H+ and FADH2 are converted to NAD+ and FAD, donating electrons and hydrogen ions to oxygen
Explanation:
<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
Answer:C The final product cannot be converted back to the original ingredients.
Answer:
VP (solution) = 171.56 mmHg
Explanation:
Vapor pressure of pure solvent(P°) - Vapor pressure of solution (P') = P° . Xm
Let's replace the data:
173.11 mmHg - P' = 173.11 mmHg . Xm
Let's determine the Xm (mole fraction for solute)
Mole fraction for solute = Moles of solute / Total moles
Total moles = Moles of solute + moles of solvent.
Let's determine the moles
Moles of solvent → 623.4 g / 119.4 g/mol = 5.22 moles
Moles of solute → 9.322 g / 180.1 g/mol = 0.052 moles
Total moles = 0.052 + 5.22 = 5.272 moles
Xm = 0.052 moles / 5.272 moles = 0.009 → 9/1000
173.11 mmHg - P' = 173.11 mmHg . 9/1000
P' = - (173.11 mmHg . 9/1000 - 173.11 mmHg)
P' = 171.56 mmHg
