Considering the distal convoluted tubule of the amphibian nephron, explain how changes in the permeability of the tubule wall to water affect the amount of pure, osmotically free water excreted in the urine. define what is meant by pure, osmotically free water.?

Osmotically free water is H2O in the urine that is not required to be there for solutes to be excreted. The highest concentration of solutes possible in amphibian urine is a collective concentration that renders the urine isosmotic to the blood plasma. This means that, for an amphibian to excrete solutes, the urine is required to contain at least an amount of H2O sufficient for the urine to be isosmotic to plasma. This amount of H2O is the amount required for solutes to be excreted. Any H2O excreted in excess of this amount constitutes osmotically free water. In the amphibian distal convoluted tubule, the permeability of the tubule wall to water is modulated to determine the amount of H2O relative to solutes in the tubular fluid. If the tubule wall is made very permeable to water, the tubular fluid comes to osmotic equilibrium with the blood plasma, meaning that the tubular fluid contains only the amount of H2O required for solute excretion and, therefore, no osmotically free water. If the tubule wall is made poorly permeable to water, the tubular fluid can be rendered very dilute relative to the blood plasma. In that case the ratio of H2O to solute in the tubular fluid can be very high, the urine may consequently contain much more H2O than required to excrete the solutes, and a large quantity of osmotically free water may be excreted.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the dissolution of polymer can be thermodynamically described in 2 steps.