These are electron dot diagrams.
They use dots arount the symbol of an atom to represent its <em>valence electron</em>s.
You will also see them called <em>Lewis dot symbols</em> or <em>Lewis diagrams</em>.
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Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Answer:
See Explanation
Explanation:
Let us consider the ionization of these compounds;
H3PO4 ⇔3H^+ + PO4^3-
H3BO3 ⇔3H^+ + BO3^3-
The next to consider is the type of electrolyte the both solution are; the both solutions are weak electrolytes and weak electrolytes do not ionize to a large extent.
The implication of this is that, not so much number of ions is added to the solution due to the poor ionization of these weak electrolytes. Hence, in spite of the subscript of 3, the conductivity of the solution does not significantly improve for the reason stated here quite unlike when strong electrolytes are used.
Answer:
There is two common methods to separate a phosphorylated protein from the proteins who does not have the phosphate group.
1. Western Blot test
The Western blot test is a method where we separate of the biological sample with SDS- PAGE (electrophoresis method) and then we transfer to a nitrocellulose membrane. We use a phospho-specific antibody to identify the protein of interest.
2. The ELISA test
this test is also a powerful method for measuring protein phosphorylation. Elisa test is more are quantitative than Western blot test and also shows great advantage in studies related with kinase activity. In this test we use a capture antibody specific for the desired protein, independent of the phosphorylation state. The target protein is binded to the antibody-coated plate. and finally a detection antibody specific for the phosphorylation site analyzed is added. In these test it is possible to use colorimetric or fluorometric detection.