As you proceed down the periodic table, the metallic character becomes stronger. This is because as the atomic radius increases, there is less attraction between the nucleus and the valence electrons due to the greater distance between them, making electrons simpler to shed.
Answer:
b-acting as a standard unit of measure
Explanation:
Question:
Which of the following statements correctly describe(s) the driving forces for diffusion of Na+ and K+ ions through their respective channels? Select all that apply.
A)The diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the plasma membrane.
B)The diffusion of Na+ ions into the cell is impeded by the electrical gradient across the plasma membrane.
C)The diffusion of K+ ions out of the cell is impeded by the K+ concentration gradient across the plasma membrane.
D)The diffusion of K+ ions out of the cell is impeded by the electrical gradient across the plasma membrane. The electrochemical gradient is larger for Na+ than for K+.
Answer:
"The concentration gradient and the electro-chemical gradient" describes the driving forces for diffusion of Na+ and K+ ions through their respective channels
Explanation:
The Na ions diffusion inside the cell is facilitated by the concentration gradient of the Na ions which is present across the plasma membrane. Hence, the diffusion of the K ions which is present outside the cell and will be impeded due to the electrical gradient which is present near the plasma membrane. Thus, the electro-chemical gradient is greater as compared to the Na ion than that of the K ion.
4 and 2 electrons are present.
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
