Answer:
1) HCl contains the Cl^- which is a good nucleophile
2) 2-methyl-2- heptanol > 2-heptanol > 1-heptanol
3) see image attached
Explanation:
If the dehydration of alcohols is carried out using HCl, the chloride ion which is a good nucleophile will attack the substrate to yield an undesirable product.
The dehydration of alcohols is an E1 reaction. Recall that the ease of E1 reaction increases in the order 3°> 2°> 1°. Hence, 2-methyl-2- heptanol forms a tertiary carbocation intermediate during dehydration and has the greatest ease of dehydration.
The three products formed during the dehydration of 3,3-dimethyl-2-butanol are shown in the image attached. Two out of the three are formed by rearrangement reactions.
C because the more amounts of mass an object has will give it a bigger gravitational field
0.3147 concentration (in moles/l) of a saline (NaCl) solution will provide an isotonic eyedrop solution.
Isotonic eye drops
Because it might result in eye discomfort or tissue damage if it is not maintained, isotonicity is regarded as a crucial component of ophthalmic medicines. A few drops of blood are mixed with the test preparation before being examined and judged under a microscope at a magnification of 40. Isotonic solutions are those that have the same amount of water and other solutes in them as the cytoplasm of a cell. Since there is no net gain or loss of water, placing cells in an isotonic solution will not cause them to either shrink or swell.
We can calculate the osmotic pressure exerted by a solution using the following expression.
π = M . R . T
where,
π is the osmotic pressure
M is the molar concentration of the solution
R is the ideal gas constant
T is the absolute temperature
The absolute temperature is 37 + 273 = 310 K
π = M . R . T
8 = (X mol/L) . (0.082atm.L/mol.K) . 310 K = 0.3147 mol/L
To learn more about osmotic pressure refer:
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Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.