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balandron [24]
4 years ago
10

Proton-alpha: "One time I lived in an environment where next door lived identical twins, Proton-beta-1 and Proton-beta-2. Boy, d

id they wreck havoc on my signal when I got zapped and had a spin flip - THEY SPLIT MY SIGNAL!!!" C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"? D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal? E) How many bonds connect these "splitting next door neighbors"? Proton-alpha: "I got them back because while they were splitting my signal, I split theirs!" F) What is the multiplicity of the Proton-betas' signal?
Chemistry
1 answer:
Mashcka [7]4 years ago
7 0

Answer:

Explanation:

C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"?

Based on n+1 rule. Here n=2 (identical beta protons).

2+1=3

So the multiplicity of alpha proton is triplet, .

D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal?

    The meaning of the adjective is this: the multiplicity of beta protons is singlet only (no spliting) in absence of alpha proton . But beta protons splits as doublet (n=1) in the presence of alpha proton,

E) How many bonds connect these "splitting next door neighbors"?

There are 3 bonds in between alpha and beta protons in a molecule.

F) What is the multiplicity of the Proton-betas' signal?

Following the  n+1 rule, here n=1 (1 alpha proton) so 1+1=2. Hence it is  a doublet.

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The answer to your question is

D: Unshared Pair

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for the following reaction, draw the major organic product and select the correct iupac name for the organic reactant. if there
denpristay [2]

<span>For number two, the equation is balanced. Na in the reactant side has one atom and also in the product side. O has only one atom in the product and in the reactant. There are 2H’s in the reactant and in the product side (present in H2O, the 2 in H stands for 2 atoms of H). One atom of Cl is present in the reactant and product side.</span>

 

For number three, the reactants are NaOH-sodium hydroxide and HCl-hydrogen chloride(hydrochloric acid). The products are NaCl-sodium chloride and H2O-water.

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Bases increase the concentration of hydronium ions by donating hydroxide ions to water molecules.
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Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4
Nataliya [291]

Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol  (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N  initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284

Now we can compute pOH,

pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37

pH = 14 - pOH = 14 - 3.37 = 10.62

(b) 20.60 mL HCl

mol HCl = 0.0206 L x 0.100 mol/L = 0.00206

mol  (CH3CH2)3N consumed = 0.0020 mol

This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in  the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol  - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

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