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Kruka [31]
2 years ago
11

Is air homogeneous or heterogeneous?

Chemistry
2 answers:
ivolga24 [154]2 years ago
8 0

Answer:  yes, Air is a homogeneous mixture of the gaseous substances nitrogen, oxygen, and smaller amounts of other substances. Salt, sugar, and substances dissolve in water to form homogeneous mixtures. A homogeneous mixture in which there is both a solute and solvent present is also a solution

Explanation:

Mademuasel [1]2 years ago
6 0

<u>pls follow me </u>

Explanation:

 a mixture is a material made up of two or more different Chemical substance/substances which are not chemically combined.A mixture is the physical combination of two or more substances in which the identities are retained and are mixed in the form of solutions, suspensions and colloids.Mixtures are one product of mechanically blending or mixing chemical substances such as elements and compounds, without chemical bonding or other chemical change, so that each ingredient substance retains its own chemical properties and makeup. Despite the fact that there are no chemical changes to its constituents, the physical properties of a mixture, such as its melting point, may differ from those of the components. Some mixtures can be separated into their components by using physical means. Azeotropes are one kind of mixture that usually poses considerable difficulties regarding the separation processes required to obtain their constituents

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What is the percent by volume of 25mL ethanol in 150 mL of water
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6 0
3 years ago
The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-} ...... (1)

At cathode (Reduction) : Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

              3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)  ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

      Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)          

         

6 0
3 years ago
what is the structure of XeO2F2???i know its hybridization as sp3d but cannot quite understand its diagram in free space,help an
skelet666 [1.2K]
The shape of XeO₂F₂ is Trigonal bi-pyramidal see-saw tetrahedron (see attached pictures)
- As you said the hybridization of Xe here is sp³d so its geometry has to be Trigonal bi-pyramidal in which F atom located on axial positions but for the final shape we exclude lone pair on Xe to give see-saw shape (see second picture) 
- Remember that we have 5 pairs (4 bond pairs + 1 lone pair) and we have to place lone pair at equatorial position.

6 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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