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AysviL [449]
2 years ago
9

In which pair is each substance a mixture?

Chemistry
1 answer:
qaws [65]2 years ago
6 0

Answer:

Air & Water

Explanation:

Air and water is the common mixtures in the book of the book called "Science & Land"

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Morgan wants to compare the weather conditions at her school each day during the week. She decides to measure and record weather
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Answer: Air Temperature

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3 years ago
What does the symbol "E3" represent?
Anna71 [15]

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your answer to your question is B

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A car was at rest. It then started moving forward in a straight line. The table shows the change in velocity.
Keith_Richards [23]

Answer:

5 m/s 2

Explanation:

if the car moved from 00 to 1-5 then to 2_10 you can know that it is going up in velocity but 5 a second so 5 m a second. hope this helps

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2 years ago
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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).

5 0
3 years ago
Answers for both boxes please ​
Elan Coil [88]

Explanation:

\frac{6.900 \times  {10}^{10} }{4.000 \times  {10}^{8} }  \\ 1.725 \times  {10}^{10 - 8} \\ 1.725 \times  {10}^{2}

Hope it will help you :)

3 0
2 years ago
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