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Vsevolod [243]
3 years ago
5

Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa

turated with oxygen at a partial pressure of 22 mm Hg and that the Hill coefficient is 3.
Chemistry
1 answer:
kumpel [21]3 years ago
4 0

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

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The substances below are listed by increasing specific heat capacity value. Starting at 30 Celsius, they absorb 100 kJ of therma
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Answer:

Silver.

Explanation:

To obtain the right answer to question, let us calculate the change in temperature for each substance assuming they all have the same mass as 100g.

This is illustrated below:

1. For Siver:

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Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.239)

ΔT = 4184°C

2. For Aluminium:

Mass (M) = 100g

Specific heat capacity (C) = 0.921J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.921)

ΔT = 1086°C

3. For Lithium:

Mass (M) = 100g

Specific heat capacity (C) = 3.56J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 3.56 )

ΔT = 281°C

4. For water:

Mass (M) = 100g

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 4.184)

ΔT = 239°C

Summary

Temperature change of each substance is given below

1. Silver => 4184°C

2. Aluminum => 1086°C

3. Lithium => 281°C

4. Water => 239°C

From the calculations made above, Silver has the highest rise in temperature.

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