What is the most common isotope of lithium?

Phenols do not exhibit the same pka values as other alcohols; they are generally more acidic. Using the knowledge that hydrogen

nevsk [136]

Answer:

Phenols do not exhibit the same pka values as other alcohols;

They are generally more acidic.

Using the knowledge that hydrogen acidity is directly related to the stability of the anion formed, explain why phenol is more acidic than cyclohexane.

Explanation:

According to Bromsted=Lowry acid-base theory,

an acid is a substance that can release $H^{+}$ ions when dissolved in water.

So, acid is a proton donor.

If the conjugate base of an acid is more stable then, that acid is a strong acid.

In the case of phenol,

the phenoxide ion formed is stabilized by resonance.

$C_6H_5OH -> C_6H_5O^- +H^+$

The resonance in phenoxide ion is shown below:

Whereas in the case of cyclohexanol resonance is not possible.

So, cyclohexanol is a weak acid compared to phenol.

7 0

3 years ago

Which group or groups of atoms are the only atoms with f orbitals A. All atoms heavier than barium B. All atoms heavier than kry

Mashcka [7]

Answer:

All atoms heavier than barium

Explanation:

In the periodic table, elements are divided into blocks. We have the;

s- block elements

p- block elements

d- block elements

f- block elements

However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.

Hence elements heavier than barium all possess f-orbitals.

6 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.