A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached Ne
1 answer:
Answer:
The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.
Explanation:
The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.
pH = -log ([H⁺])
American Midwest
pH = -log ([H⁺])
5.02 = -log ([H⁺])
[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M
New England
pH = -log ([H⁺])
4.66 = -log ([H⁺])
[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M
The ratio of concentrations is:
The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.
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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).
