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Anestetic [448]
3 years ago
10

All single-displacement reactions can be classified as another type of reaction as well. What type of reaction is that?

Chemistry
2 answers:
user100 [1]3 years ago
8 0

combustion because one atom in wrong place will make it combust

Lyrx [107]3 years ago
5 0

Answer:

The single displacement reaction can be classified as <u><em>redox</em></u>

Explanation:

A simple substitution reaction or simple displacement reaction, called single-displacement reaction, is a reaction in which an element of a compound is substituted by another element involved in the reaction. The starting materials are always pure elements and an aqueous compound. And a new pure aqueous compound and a different pure element are generated as products. The general form of a simple substitution reaction is:

AB + C ⇒ A +BC

where C and A are pure elements; C replaces A within compound AB to form a new co, placed CB and elementary A.

A single-displacement reaction is characterized by a place of exchange of cations or anions with another to form a new product. A begins as an ion in the solution and appears elementally in the reaction products. In case of reagent C it starts in its elemental form and ends as an ion in an aqueous state forming compound BC.

On the other hand, oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The redox reaction is a single-displacement reaction.  A chemical substance goes from being an ion to a pure element. The oxidation state of a monoatomic ion in solution is equal to its charge, while the oxidation state of a pure element is equal to zero. Any element that goes from being an ion to a pure element or vice versa must be changing oxidation state.

A displaces C, which can be the H₂ molecule or another metal. If C is formed by the reduction of ions, then species A, by the conservation of mass and energy and must provide electrons: it must oxidize.

On the other hand, if A and C are metallic species, but C is in the ionic form (M+) and A in its natural state, then the displacement reaction will occur only if A is more active than C, forcing the latter to Accept the electrons to reduce to their metallic state (M).

So finally, <em><u>the single displacement reaction can be classified as redox</u></em>.

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5 0
3 years ago
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
How many electrons will each element gain or lose in forming an ion?
Amanda [17]
It depends on the number of valence electrons required to make octet or duplet( in case of H). For example, Nitrogen(atomic number = 7) has electronic configuration(2,5) which means nitrogen has 5 valence electrons and requires 3 more electrons to complete its octet. After gaining 3 electrons from atoms of an element with less electronegativity than N, it forms nitride ion (N ^{-3}). Hope this helps.
8 0
3 years ago
For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this
Salsk061 [2.6K]

The following are the steps  to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3                           N = +5                        

O = -2                            O = -2

Ag = +1                         Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

     NO2– -----> NO3–

     Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

     NO2– + H2O -----> NO3–

     Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

    NO2– + H2O -----> NO3– + 2H+

     Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     2 Ag+ + 2e------> 2 Ag(s)

3 0
3 years ago
The effusion rate of hcl is 43.2 cm/min in a certain effusion apparatus. what is the rate of effusion of ammonia in the same app
Jlenok [28]

The effusion rate is 1.125 cm/sec for ammonia.

How to find effusion rate ?

Effusion rate (r1) HCl = 43.2 cm/min

Molar mass (m2) NH3 =17.04g/mole

Molar mass (m1)  HCl    =36.46g/mole

  • Substitute the molar masses of the gases into Graham's law and solve for the ratio.
  • r1÷r2=√m2÷m1

       firstly convert 43.2 cm/min into cm/sec i.e., 0.72 cm/sec

      Then,

      0.72/r2 =√17.04/36.46

      r2= 1.125 cm/sec

Hence, the rate of diffusion of ammonia is 1.125 times faster than the rate of diffusion of hydrogen chloride.

learn more about effusion here:

brainly.com/question/2097955

#SPJ4

5 0
1 year ago
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