Which molecule plays the greatest role in the thermal regulation of the troposphere?
1 answer:
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Answer:
look at the graph
Explanation:
We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.
Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.
So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)
Answer:
An example of oxygen–hemoglobin (O2–Hb) dissociation curves from (A) one penguin at pH 7.5, 7.4 and 7.3, and (B) the emperor penguin, the bar-headed goose (Anser indicus) (Black and Tenney, 1980) and the domestic duck (Anas platyrhynchos, forma domestica) (Hudson and Jones, 1986) at pH 7.4. Note that as for the bar-headed goose, the O2–Hb dissociation curve of the emperor penguin is significantly left-shifted as compared with the domestic duck (and most birds). The bar-headed goose photo is courtesy of Graham Scott; the domestic duck photo is by Maren Winter (licensed under the terms of the GNU Free Documentation License, Version 1.2 or any later version); the penguin photo is by J.M.
Explanation:
The resulting regression equations from the plots of log[SO2/(100–SO2)] vs log(PO2) (all saturation points, all penguins combined) were:
pH 7.5: log[SO2/(100–SO2)] = 2.92589 × log(PO2) – 4.24338 (N=43, r2=0.98, P<0.0001),
pH 7.4: log[SO2/(100–SO2)] = 2.94767 × log(PO2) – 4.39858 (N=70, r2=0.98, P<0.0001),
pH 7.3: log[SO2/(100–SO2)] = 3.04945 × log(PO2) – 4.72019 (N=38, r2=0.99, P<0.0001),
pH 7.2: log[SO2/(100–SO2)] = 3.15958 × log(PO2) – 4.97618 (N=9, r2=0.99, P<0.0001).
