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gayaneshka [121]
3 years ago
12

4. You are studying the behavior of a perfluorocarbon-based O2 carrier as an emergency treatment for hypoxemia. It has an approx

imate molecular radius of 2 nm. Consider ultrafiltration through the kidney (glomerular membrane pore radius 4 nm, length 800 nm). Assuming a therapeutic dose in the blood of 22 mM and 2 mM in the filtrate, determine the electrochemical potential across glomerular membrane.
Chemistry
1 answer:
Levart [38]3 years ago
7 0

Answer:

See explaination

Explanation:

The molecular radius of the perflurocarbon-based O2 carrier suggests that it passes through the Bowman’s capsule during ultrafiltration. Initial levels of the O2 carrier in blood are 22mM and concentration of the O2 carrier in the final glomerular filtrate is 2mM. Electrochemical potential is the driving source of energy required for the transport of a solute across the membrane. It is also known as the Gibbs free energy G. ΔG’o is the free energy change under standard conditions. Standard free energy change is directly related to the equilibrium constant (K’eq). Equilibrium is the state of the reaction where the concentration of the reactants equals the concentration of the products.

K’eq = [final concentration]/[initial concentration]

K’eq = 2mM/22mM = 0.09

ΔG’o = -RT ln K’eq ; R (gas constant) = 8.315 J/mol.K

T (temperature) = 298 K

ΔG’o = - (8.315 x 298) ln 0.09 = +5.96 kJ/mol

The separation of small molecules and ions from large molecules and cells in the blood is termed as ultrafiltration. The filtered out fluid is known as glomerular filtrate, capsular filtrate or ultrafiltrate. A nephron is a single unit of structure and function in a kidney.

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The equation for the formation of water from hydrogen gas and oxygen gas is 2H2 +
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Answer:

72g

Explanation:

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Which of the following statements is true?
Lera25 [3.4K]

The  statement  which is true is

metals  lose  electrons  to become cations

<u><em>Explanation</em></u>

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3 0
3 years ago
Read 2 more answers
C12H22011+1202--&gt;12CO2+11H20
kogti [31]

Answer:

0.185moles

Explanation:

Given parameters:

Volume of O₂  = 49.8L

Unknown:

Number of moles of sucrose required  = ?

Solution:

We can assume that the reaction takes place at standard temperature and pressure.

From this, we can find the number of moles of oxygen that reacted and extrapolate to that of sucrose.

Chemical equation;

           C₁₂H₂₂0₁₁  +  120₂   →   12CO₂   +   11H₂0

Number moles  = \frac{volume of gas}{22.4}   at STP

Number of moles of oxygen gas = \frac{49.8}{22.4}   = 2.22moles

           12 moles of oxygen gas combines with  1 mole of sucrose

       2.22 moles of oxygen gas will combine with \frac{2.22}{12}   = 0.185moles

4 0
4 years ago
An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down
jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0
3 years ago
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