Warm air from above the water rises and is replaced by the cool air from the shore
Answer:
CH2
Explanation:
We are given that a compound is 14.40% hydrogen and 85.60% carbon.
Let the mass of the substance be 100g.
Mass of the hydrogen: 14.40% of 100 = 14.40 g
Mass of the carbon: 85.60% of 100 = 85.60 g
Now, let's find the moles of hydrogen:

Moles of carbon:

Let's put these in a ratio and simplify:
7.13 mole C: 14.40 mole H
1 mole C: 2 mole H
Therefore, the empirical formula of this compound is CH2.
Hope this helps!! If you have any questions about my work, please let me know in the comments!
The answer is each indicator has a narrow range. We need many different indicators to span the entire ph spectrum because each indicator has a narrow range.
Answer:
4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.
Explanation:
The cytochromes are <u>proteins that contain heme prosthetic groups</u>. Cytochromes <u>undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome</u>:

The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes <em>b</em> and <em>c₁</em>, and a Fe-S center. This second complex reduces cytochrome <em>c</em>, a water-soluble membrane peripheral protein. Cytochrome <em>c</em>, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes <em>a</em>, <em>a₃</em> and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.
Each transporter "downstream" is <u>more electronegative</u><u> than its neighbor </u>"upstream"; oxygen is located in the inferior part of the chain. Thus, the <u>electrons fall in an energetic gradient</u> in the electron chain transport to a more stable localization in the <u>electronegative oxygen atom</u>.