Which sentence uses the positive form of an adverb?

What does the process of natural selection involve

Lorico [155]

Answer:

Only the organisms best adapted to their environment tend to survive and transmit their genetic characters in increasing numbers to succeeding generations while those less adapted tend to be eliminated.

Explanation:

According to Darwin's Theory of Evolution.

Hope this helped.

5 0

3 years ago

Explain the volume of ideal gasses as related to the molar concept.

horrorfan [7]

Answer:

n = Initial volume/22.4L

Explanation:

The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.

Now, in terms of molar mass, number of moles is given by the formula;

n = mass of the sample/molar mass

In terms of avogadro's number, number of moles is;

1 mole = avogadro's number = 6.02 × 10^(23)

Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.

Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.

Thus;

n = Initial volume/22.4L

6 0

3 years ago

Describe how you would crystallise potassium nitrate from its aqueous liquid?

Rus_ich [418]

Just look here https://www.jiskha.com/search/index.cgi?query=describe+how+you+can+crystallize+potassium+nitrate+fro...

5 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

How many electrons are found in their outer shell ?

dezoksy [38]

The answer is eight electrons.

3 0

3 years ago