Answer:
Major Product = 4-chloro-4-methylcyclohex-1-ene
Explanation:
Alkene are the class of organic compounds which contain one or more double bonds between two carbon atoms. Alkenes are considered most reactive among the unsaturated hydrocarbons and they undergo <em>addition reactions</em> due to high electron density around the double bonds.
In given question it is written that we are provided with one equivalent of HCl while, our compound contains two double bonds (diene) so in selected starting material the HCl will be added across (hydrohalogenation reaction) the substituted double bond because it will give a more stable carbocation (<u><em>tertiary carbocation</em></u>) during the reaction course. Hence, as shown in reaction scheme 4-chloro-4-methylcyclohex-1-ene will be the major product.
How does it what. i don’t know if there’s a photo but can’t see it
Answer:
Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.
After that if we to multiply the avogadro number with it. So 3 *NA
Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.
So final claculation is =3*2*NA.
Explanation:
your welcome
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Answer:
Explanation:
A) False.
Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.
Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.
B) True.
Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.
C) True.
Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.
D) Wrong.
Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.
E) Wrong.
The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.
Answer:
The concentration of KBr is 
Explanation:
From the question we are told that
The mass of KBr is 
The molar mass of KBr is 
Volume of water is 
This implies that the volume of the solution is 
The number of moles of KBr is
Substituting values
The concentration of KBr is mathematically represented as
