Answer:
C Region
Explanation:
C Region contains all the liquids as 0 °C is the freezing point of water (Crystals of water are formed leaving it no more in the liquid state) and 100 °C is the boiling point (The water boils leaving it no more in the liquid state).
Hence, All liquids are contained in the C region.
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Hope this helped!
<h3>~AH1807</h3>
Moles = 15.5 g / 40 g/mol = 0.3875 mol
M = 0.3875 mol / 0.250 L = 1.55M
Answer: The answer to this question is transpiration.
Explanation: I know this answer because i looked it up in a book. The other explanation is I study about this a lot.
Answer:
is usually structurally similar to the substrate.
Explanation:
Competitive inhibitors resemble normal substrate and binds to enzyme at the active site usually and prevents substrate from binding.
Active sites are main location for the substrate-enzyme binding. These sites involve weak as well as reversible bonds between the substrate and the enzyme. These inhibitors bind to the active sites and form weak and reversible bonds. Competitive inhibitors can be dissociated from active site by increasing concentration of the substrates. Substrates has to compete for active site and displace the bound competitive inhibitors.
<u>Hence, correct option is - is usually structurally similar to the substrate.</u>
Answer:
Structure in attachment.
Explanation:
The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.
In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.