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cluponka [151]
3 years ago
8

Tiffany was investing how fast it took Hayden to react to different sounds. Identify the dependent variable.

Chemistry
2 answers:
klio [65]3 years ago
5 0
D because it is him who is the dependent variable
erma4kov [3.2K]3 years ago
4 0

Answer:

B

Explanation:

I think

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On the pH scale pure water has a pH of<br> A. 0<br> B. 7<br> C. 14
konstantin123 [22]

Answer:its b 7

Explanation:

7 is neutral and pure water is the middle man

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What happens to the electrons in an ionic bond and a covalent bond?
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In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.

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True or false: All heterotrophs are omnivores.
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False. They can be both omnivores and carnivores.

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In an animal, a muscle cell requires more energy than other cells. Because of this, you would
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At normal blood pH pH (7.4), hemoglobin is 80 80 % saturated at a partial pressure of oxygen ( O 2 O2 ) of 40 mmHg 40 mmHg . Use
schepotkina [342]

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).

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