1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lana [24]
3 years ago
5

Which substance in the reaction serves as the electrophile?'?

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
4 0

Electrophiles are reagents attracted to electrons.

Electrophiles tend to be electron-deficient and carry partial positive charges. They are attracted to species with lone pairs of electrons. For example, protons \text{H}^+ have no electrons and tend to share ones with other species, hence behaving as electrophiles in aqueous reactions. In the reaction between \text{H}^+ and ammonia \text{NH}_3, protons would be attracted to lone electron pairs on nitrogen atoms in ammonia molecules, which carry partial positive charges.

The Lewis Acid-base theory define Acids as species that accept electron pairs in a particular acid-base reaction. Electrophiles, by definition, tend to accept electrons. Lewis acids thus behaves as electrophiles in acid-base reactions. In the previous example, \text{H}^+ demonstrates acidic behavior and can be inferred as an electrophile.

You might be interested in
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
4 years ago
What are the properties of elements related to?
balu736 [363]
The answer you are looking for is A. the average atomic mass hope this helped have a nice day :)
3 0
3 years ago
Produced by the stomach to destroy bacteria
dalvyx [7]

Answer:

Your stomach lining also secretes hydrochloric acid, which creates the ideal conditions for the protein-digesting enzymes to work. The potent hydrochloric acid kills bacteria, protecting your body from harmful microbes which can enter your body in food.

4 0
3 years ago
Drag each tile to the correct box.
77julia77 [94]

this is my attachment answer hope it's helpful to you

3 0
3 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
Other questions:
  • Need help in chemistry with these 10 questions please. 80 points well be given.
    8·2 answers
  • The most recent treason indictment involved a U.S. citizen who was accused of turning into an operative for which of the followi
    15·1 answer
  • The enegry need to pump sodium outside the cell in active transport is
    13·1 answer
  • What are some examples of how electrolysis is used in scientific research ?
    8·1 answer
  • I NEED HELP PLEASE, THANKS! BRAINLIEST :)
    5·2 answers
  • What structures are found in<br> plant cells that are not found in animal cells?
    5·1 answer
  • 3.Write the chemical equation for the reaction when methane burns in (2)
    7·1 answer
  • Why is it difficult to detect the presence of isotopes of an element during chemical<br> studies?
    14·1 answer
  • "Working on rigs is dangerous due to explosions and weather"
    6·2 answers
  • Defination of dative covalent bond
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!