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RSB [31]
1 year ago
8

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

items as equivalent, overlap them. resethelp hardesteasiest

Chemistry
1 answer:
marysya [2.9K]1 year ago
3 0
The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
                                                              The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

                                               B   >  C  >  A

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How many atoms are in a molecule of RSq?
Wewaii [24]
2 atoms in a molecule of RSq, TSq contains 3 different types of atoms (Sq, R, and T.)
6 0
2 years ago
Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57
vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0
1 year ago
Consider the compounds 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. using silica gel plates, which of the two compoun
grandymaker [24]
1- we know that 4-tert-butylcyclohexanol is more polar than 4-tert-butylcyclohexanone (where the alcohols in general are more polar than ketons due to the hydrogen bond)
2- during separation via chromatography (in this case) the more polar solute will dissolve easily in polar solvents, where like dissolves like.
3- So, 4-tert-butylcyclohexanol will dissolve in ethyl acetete (which is polar) more than 4-tert-butylcyclohexanone, i.e, will have much higher Rf.
4- And also 4-tert-butylcyclohexanone will dissolve in dichloromethane (which lower in polarity than ethyl acetate) more than 4-tert-butylhexanol, i.e, will have much higher Rf 
3 0
2 years ago
Imagine you are a detective examining a crime scene. You are trying to
Sidana [21]

Answer:

C. pieces of hair found at the crime scene.

Explanation:

using the pieces of hair, you can get the DNA of the person. This will give u a better lead in solving the crime.

Hope it helps u ....

4 0
2 years ago
Read 2 more answers
An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered
Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0
1 year ago
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