All categories

3 years ago

CAN HF USED TO CLEAVE ETHERS EXPLAIN

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

Explanation:

Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.

You might be interested in

Suppose that NaCl is added to hexane (C6H14) instead of water. Which of the following intermolecular forces will exist in the sy

svet-max [94.6K]

Answer:

Ion-ion force between Na+ and Cl− ions

London dispersion force between two hexane molecules

Explanation:

"Ion-dipole force between Na+ ions and a hexane molecule " does not exist since hexane has only non-polar bonds and therefore no dipole.

"Ion-ion force between Na+ and Cl− ions " exists since both are ions.

"Dipole-dipole force between two hexane molecules " does not exist since hexane molecules do not have a dipole.

"Hydrogen bonding between Na+ ions and a hexane molecule " does not exist since the hydrogen in the hydrogen bond must be bonded directly to an electronegative atom, which hexane does not have since it is a hydrocarbon.

"London dispersion force between two hexane molecules" exist since hexane is a molecular compound.

5 0

3 years ago

Read 2 more answers

NOOO LINNKKKSSS

Elena L [17]

The answer is D I hope this helps you !

3 0

3 years ago

Read 2 more answers

Water is Carried to the leaves by stomata

tiny-mole [99]

Answer:

please explain further and i maybe can help you

Explanation:

5 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago