Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH will be,
The intermediate balanced chemical reaction will be,
..[1]
..[2]
..[3]
Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:
We get :
..[1]
..[2]
[3]
The expression for enthalpy of formation of 
will be,
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Four people weigh a standard mass of 10.00 g on the same balance. The set of readings suggest measurements that are neither precise <span>nor accurate is the one with less mass</span>
Answer:
Explanation:
An atom is the smallest unit of an element that can take part in a chemical reaction. Atoms are made up of protons, neutrons and electrons. Atoms can exist as a monoatomic (such as in the case of Helium, Xenon and Neon) or as diatomic (such as in the case of oxygen and nitrogen). Atoms take part in a chemical reaction and there reactivity varies among themselves.
From the above, it can be deduced that atoms have protons, neutrons and electrons. The number of protons (which is positively charged) of an atom determines it's position on the periodic table because elements in the periodic table are arranged according to the number of protons (called atomic number). The electron(s) present in the outermost shell of each atom (called valence electrons) determines there chemical reactivity. What happens here is that, all atoms (except noble gases) want to achieve there duplet or octet configuration so as to become stable. This octet configuration means they want to have there outermost shell completely filled (with eight electrons or two electrons for duplet). They usually achieve this configuration by taking part in chemical reactions. Thus, when an atom has just one electron in it's outermost shell, it becomes easy to lose it to another atom by way of interacting with it in a chemical reaction. When it loses this single electron (valence electron) in it's outermost shell, it becomes stable with the inner completely filled shell (that would be the new outermost shell). Examples include Lithium, sodium and potassium. Sodium (with eleven electrons and three shells) would lose the single electron in it's outermost shell so as to have just two shells with the second shell completely filled with eight electrons. Thus, <u>the more the valence electron to be lost to achieve the octet structure</u>,<u> the lesser the reactivity of the atom</u>.
Also, an atom that has just one electron to complete it's own outermost shell and thus achieve it's octet structure is also highly reactive. This is also because it is easy for this atom to receive a single electron and become completely filled. Examples include chlorine, fluorine and iodine. Fluorine (with nine electrons and two shells) will easily accept one more electron so as to achieve it's octet structure with a completely filled outermost shell (of eight electrons). Thus, <u>the lesser the electrons to be gained to achieve the octet configuration, the higher the chemical reactivity of such atoms</u>. Noble gases have extremely low or no reactivity at all for this reason because it has a completely filled outermost shell (no losing or donating).
It should also be noted that metals (which are found on the left of the periodic table) exist as monoatomic while gases (which are found on the right), with the exception of noble gases, are mostly diatomic.
Answer: An increase in the ratio of insulin to glucagon will increase the activity of --
- Acetyl-CoA carboxylase(+)
-Phosphofructokinase PFK2(+)
-Glycogen synthase(+)
- Hormone sensitive lipase (-). The hormone sensitive lipase activity is not increased with increased insulin activity.
Explanation: increased insulin - glucagon ratio is usually high in fed state.Insulin helps the cells absorb glucose, reducing blood sugar and providing the cells with glucose for energy. When blood sugar levels are too low, the pancreas releases glucagon. Glucagon instructs the liver to release stored glucose, which causes blood sugar to rise.
