Question

-Kor Cor The Henry's law constant of oxygen in water at 25 °C is 773 atm mol'' kg of water. Calculate the molality of oxygen in water under a partial pressure of 0.20 atm. Assume that the solubility of oxygen in blood at 37 °C is roughly the same as that in water at 25 °C, comment on the prospect for our survival without hemoglobin molecules. The total volume of blood in the human body is about 5 L The potential difference across a cell membrane is found to be 40 mV, with the inside negative to the outside. Assume an equilibrium distribution, and calculate the concentration ratio CMg(out)/CMe(in) for Mg2+ ions; T = 37 °C. 10. A solution of tRNA at a concentration of 100 mg/ml is dialyzed to equilibrium against 0.1 M NaCl, Calculate the Donnan potential and the concentration ratio CNatDNA solution)/CNa(dialysate). Assume there is one charge per nucleotide of M=340(molecular wt); T = 37 °C. According to the chemiosmotic theory, an electrochemical proton gradient is used to synthesize ATP in mitochondria. The oxidation of carbohydrates and fats is used to pump protons outside the mitochondrial membrane until the steady state membrane potential is -140 mV and the pH gradient is ApH=1.5. Inside the mitochondrion, pH=7,0, (ATP] = 1 mM, [Pi] = 2.5 mM, (ADP) = 1 mM, and T = 298 K. ADP + P === ATP + H20 L AG° = 31.0 kJ/mol How much chemical potential is required to synthesize ATP inside the mitochondria? GENER Tela How much free energy is made available by moving I mole of H* from the

NUMBER 10 PLEASE

Answer 1

8) assuming the same solubility in blood, this is very low amount of oxygen carried by 1 litre of blood .So 5 litre of blood carry 1.290 mMol oxygen .

..Griviem..Kpa.(partial pressure 04.02.) – 0..20. atm... OR=Kpa..20. kPa... .. Henry's law constant for 2 - 773(atm / mall likes 20. Molar Solubility 10.02.in water - Suvery of Henry's law.constant & Q_ partial presure of gas ....0...20.. 773 ...0.00129.......20 .........0..0.0.0.298.mal. /liter .OR.........258.m. Mol. liter Oz dissalved. . en water ..

Now considering the total Hb in 1 litre of blood then ..

13.5 to 17.5 gm/ dl of blood carry oxygen = 1.34 (ml of O2)*17.5

=23.45 ml of oxygen carry's by 1 DL of HB

Then 1 litre of blood carrys = 23.45*10⁶

= 234.5*10⁵ ml of oxygen

So with this much amount of oxygen the solubility is much low so without haemoglobin, the survival is too difficult without haemoglobin.