2. Is it true thatOnly the stratosphere contains enough air for living things to breathe?
1 answer:
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Answer:
15.24 J is the change in internal energy for this process.
Explanation:
According to the first law of thermodynamics:-
Where,
U is the internal energy
q is the heat
w is the work done
From the question,
q = + 75 J
(+ sign as the heat is being absorbed)
The expression for the calculation of work done is shown below as:
Where, P is the pressure
is the change in volume
From the question,
= 2.00 - 1.40 L = 0.60 L
P = 1.02 atm
Also, 1 atmL = 101.3 J
So,
(work is done by the system)
So,
<u>15.24 J is the change in internal energy for this process.</u>
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).
