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MissTica
2 years ago
7

Consider the compounds 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. using silica gel plates, which of the two compoun

ds will have the higher rf value when ethyl acetate is the developing solvent? which will have the higher rf value when dichloromethane is the developing solvent? give your reasoning in each case.
Chemistry
1 answer:
grandymaker [24]2 years ago
3 0
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 
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What two-step process separated the cans into aluminum, steel, and tin?
Vikentia [17]

Answer:

Magnet

Durability and heaviness.(texture)

Explanation:

Magnet can be use to separate Aluminum from mixture of steel and aluminum.

Though aluminum and steel look alike but magnet can be use to separate it.

If the can attract the magnet or magnet stick to the can, it is a steel can. Aluminum does not stick to magnet.

A mixture of Aluminum and tin can also be separated by magnet.

Tin attract magnet but tin is more durable, heavy and does not corrode easily.

When u touch the three cans, tin is heavy and durable.

4 0
2 years ago
To prepare a 2 M solution of potassium nitrate (KNO3), which quantities must be measured? The mass of the and the volume of the
Olin [163]

Answer:

The mass of the solute and the volume of the solution.

Explanation:

Hello,

In this case, given the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.

Best regards.

4 0
2 years ago
Read 2 more answers
At this point Ron is slightly confused, this isn’t surprising. However, Hermione is doing rather well with them. This also isn’t
Zigmanuir [339]

Answer:

\boxed{\text{0.780 atm}}

Explanation:

Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.

She shows Ron how to use the Ideal Gas Law to solve the problem.

pV = nRT

She collects the data:

V = 1.00 L; n = 0.0319 mol; T = 25.0 °C

She reminds him to convert the temperature to kelvins

T = (25.0 +273.15) K = 298.15 K

Then she shows him how to do the calculation.

p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}

Isn't she smart?

4 0
2 years ago
What is the value of the σ∗1s mo wave function at the nodal plane?
Leona [35]

Answer:

There is None

Explanation:

This is because it is a derived function dependent on other factors.

5 0
1 year ago
Read 2 more answers
A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

Now put all the given values in above formula, we get:

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

Now put all the given values in above formula, we get:

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0
1 year ago
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