Answer:
Left and right: two forces
Total: four forces
Explanation:
If it's just the forces that they are using it would be two the force of each person's hand on the other. However there is also gravity pulling down on their hand but they use another force to pull it up which you can call the normal force.
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Although phlorizin inhibition of Na+-glucose cotransport occurs within a few seconds, 3H-phlorizin binding to the sodium-coupled glucose transport protein(s) requires several minutes to reach equilibrium (the fast-acting slow-binding paradigm). Using kinetic models of arbitrary dimension that can be reduced to a two-state diagram according to Cha’s formalism, we show that three basic mechanisms of inhibitor binding can be identified whereby the inhibitor binding step either (A) represents, (B) precedes, or (C) follows the rate-limiting step in a binding reaction. We demonstrate that each of mechanisms A–C is associated with a set of unique kinetic properties, and that the time scale over which one may expect to observe mechanism C is conditioned by the turnover number of the catalytic cycle. In contrast, mechanisms A and B may be relevant to either fast-acting or slow-binding inhibitors.
Explanation:
Answer:
Option d.
Explanation:
Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:
R-C(=O)-R'
The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).
The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.
Hence, from all of the above <u>we can discard the option a</u>, (the carbonyl groups exhibits sp2 hybridization), <u>the option b</u> (carbon-oxygen bond is a bond polar) and <u>the option c</u> (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).
Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.
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