When the pH of a solution is 12.83, what is [H +]?

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

marysya [2.9K]

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

3 0

3 years ago

What is the pH of a solution with a 1.50x10-9 M hydroxide ion concentration?

dimaraw [331]

Answer: 5.18

Explanation:

Mathematically, pOH is expressed as

pH = -log(OH-)

where OH-is the concentration of hydroxide ion

So, pOH calculations are as follows

pOH = -log(1.50x10-9 M)

pOH = -(-8.82)

pOH = 8.82 [the two minus signs cancelled out]

Since pOH = 8.82; apply the formula

pH + pOH = 14 to get pH of the solution

Hence, pH + pOH = 14

pH + 8.82 = 14

pH = 14 - 8.82

pH = 5.18

Thus, the pH of a solution with a 1.50x10-9 M hydroxide ion concentration is 5.18 (slightly acidic)

4 0

4 years ago

The first step in the formation of all three products is the loss of the ots leaving group to give a carbocation intermediate. t

lina2011 [118]

The carbocation stabilized by resonance structure and thereby lowers the energy of the carbocation, hydrogen will add to the carbon in the double bond that produces delocalization of electrons.

<h3>What is carbocation?</h3>

A carbocation is a molecule in which a carbon atom has a positive charge and three bonds.

In general, electrons are stabilized by delocalization. The stabilization energy engendered by delocalization over more than two atoms is called the resonance stabilization energy or simply the resonance energy. The greater the extent of electron delocalization the greater the resonance stabilization.

Learn more about the carbocation here:

brainly.com/question/19168427

#SPJ1

4 0

2 years ago

Ethanol melts at -114 degree C. The enthalpy of fusion

Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

$(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ethanol = 25.0 g

$c_{p,s}$ = specific heat of solid ethanol= 0.97 J/gK

$c_{p,l}$ = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = $\frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

$T_{final}-T_{initial}=\Delta T$ = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

$\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]$

$\Delta H=6885.93J=6.88kJ$ (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0

3 years ago

The product of the nuclear reaction in which 40Ar is subjected to neutron capture followed by alpha emission is ________. The pr

Orlov [11]

Answer:

37S

Explanation:

Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.

We have two main types of radioactivity: radioactive decay and artificial transmutation.

In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.

23892U = 23490Th + 42He

Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.

147N + 42He = 178O + 11P

For the above question which is artificial transmutation, the reaction equation is

4018Ar + 10n = 3716S + 42He

So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.

5 0

3 years ago