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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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2 years ago
What name is given to a polymer where all of the substituents (i.e., the R groups) are in the same relative orientation (i.e., o
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3 years ago
4. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0˚C in a calorimeter,
myrzilka [38]

Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

Volume of 1.00 M Na2SO4 = 1.00 L

enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL

Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

Q= m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

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Answer:

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Why do hotspots appear to move in relation to tectonic plates?
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Answer:The heat that fuels the hot spot comes from very deep in the planet. This heat causes the mantle in that region to melt. The molten magma rises up and breaks through the crust to form a volcano. While the hot spot stays in one place, rooted to its deep source of heat, the tectonic plate is slowly moving above it.

Explanation:

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3 years ago
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