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4 years ago

why should the pka of a leaving groups conjugate acid be considered when performing a substitution reaction

1 answer:

7 0

<span> A : pKa values have little effect on a functional group's ability to leave. There is no need to consider this when outlining a substitution reaction. B : In order to have a substitution reaction a compound needs to a have a functional group that will serve as a good leaving group. A high pKa (pKa > 5) of a conjugate acid produces the weakest conjugate bases in a substitution reaction. C : In order to have a substitution reaction a compound needs to a have a functional group that will serve as a poor leaving group. A low pKa (pKa > 0) of a conjugate acid produces the weakest conjugate bases in a substitution reaction. D : In order to have a substitution reaction a compound needs to a have a functional group that will serve as a good leaving group. A low pKa (pKa < 0) of a conjugate acid produces the weakest conjugate bases in a substitution reaction. This favors a substitution reaction.</span>

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Hybridization _____.

zzz [600]

The answer is number two here is an explanation of why I said number two ....

In chemistry, orbital hybridisation is the concept of mixing atomic orbitals into new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory.

5 0

4 years ago

What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?

tankabanditka [31]

The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ

6 0

3 years ago

Read 2 more answers

The work function is the energy that must be supplied to cause the release of an electron from a photoelectric material. The cor

vladimir2022 [97]

Answer:

No photoelectric effect is observed for Mercury.

Explanation:

From E= hf

h= Plank's constant

f= frequency of incident light

Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23

f= 11×10^14 Hz

The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface

3 0

4 years ago

Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0

3 years ago

Airida [17]

Answer:

Final pressure = 362.7 Pa

Explanation:

Given that,

Initial volume, V₁ = 930 ml

Initial pressure P₁ = 156 Pa

Final volume, V₂ = 400 mL

We need to find the final pressure. We know that the relation between volume and pressure is inverse i.e.

$V\propto \dfrac{1}{P}\\\\\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}\\\\P_2=\dfrac{V_1P_1}{V_2}\\\\P_2=\dfrac{930\times 156}{400}\\\\P_2=362.7\ Pa$

So, the final pressure is equal to 362.7 Pa.

3 0

3 years ago