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zhannawk [14.2K]
3 years ago
8

If the human body contains about 50 deciliters of blood, and the blood has about 15 grams of hemoglobin per deciliter, and all o

f the hemoglobin molecules are saturated with O2, how many grams of O2 will the blood contain?_______________________
Chemistry
2 answers:
Tcecarenko [31]3 years ago
6 0

Answer:

The blood will contain 750 grams of O2

Explanation:

Volume of blood in the human body = 15 deciliters

Mass of hemoglobin per deciliter of blood = 15 grams

Mass of hemoglobin in 50 deciliters of blood = 50×15 = 750 grams

Since all the hemoglobin molecules are saturated with O2, mass of O2 in the blood will be the same as mass of hemoglobin molecules in the blood.

Therefore, mass of O2 in the blood is 750 grams

lubasha [3.4K]3 years ago
3 0

Answer:

750 g of O2 in the blood.

Explanation:

Volume of blood in the human body, V = 15 deciliters

Density of hemoglobin, D = 15 grams/deciliter

Mass of hemoglobin = Density * volume

= 50 * 15

= 750 g.

Since all the hemoglobin molecules are saturated with O2, mass of O2 in the blood = mass of hemoglobin molecules in the blood. Mass of O2 in the blood is 750 g.

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An equilibrium mixture of PCl₅(g), PCl₃(g), and Cl₂(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
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Answer:

PCl₅ = 223.4 torr

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For gas substances, the equilibrium constant can be calcultaed based on the partial pressures (Kp). For a generic reaction:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}, where pX is the partial pressure of X.

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Kp = 0.803

When more Cl₂ is added, for Le Chatêlier's principle, the equilibrium will shift for the left, more PCl₅ will be formed, and the equilibrium will be reestablished.

The initial total pressure was 243.4 torr, so if it jumps to 263.0 torr, it was added 19.6 torr of Cl₂, so the partial pressure of Cl₂ is 32.8 torr. For the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

217.0         13.2        32.8    <em>  Initial</em>

+x               -x            -x         <em>Reacts</em> (stoichiometry is 1:1:1)

217 + x       13.2-x     32.8-x  <em>Equilibrium</em>

So, the equilibrium constant must be:

Kp = \frac{(13.2-x)*(32.8-x)}{217+x}

0.803 = (432.96 - 46x + x²)/(217 + x)

432.96 - 46x + x² = 174.251 + 0.803x

x² - 46.803x + 258.71 = 0

By Bhaskara's equation:

Δ = (46.803)² - 4*1*258.71

Δ = 1,155.68

x =[-(- 46.803) ±√1,155.68]/2

x' = (46.803 + 33.99)/2

x' = 40.40

x'' = (46.803 - 33.99)/2

x'' = 6.40

x < 32.8, so x = 6.40

The new partial pressures are:

PCl₅ = 217.0 + 6.40 = 223.4 torr

PCl₃ = 13.2 - 6.40 = 6.8 torr

Cl₂ = 32.8 - 6.40 = 26.4 torr

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