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2 years ago

The sodium atom loses an electron to form a sodium ion (Na+). Which statement is correct with respect to its atomic radius?

2 answers:

6 0

The atomic number of sodium is 11. Sodium is a metal and it has the electronic configuration: 2, 8,1. This configuration implies that, the electrons in an atom of sodium are distributed into three different shells and the outermost shell has one electron in it. If sodium decide to give away the electron on its outermost shell, it will have 10 electrons left and those electrons will be distributed in only the first two shells and then it will be described as an ion. The third shell will not exist again because the electron there has been given away. Thus, the sodium ion is going to have a smaller atomic radius because its size has been reduced. This implies that the sodium ion will have a smaller radius than the sodium atom.

mixer [17]2 years ago

3 0

Its atomic radius increases because there is less electron shielding after the sodium loses one e-. Sodium is a nonmetal which means it forms an anion. Anions always increase from their original atomic radius

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How many grams are in 0.36 moles of CCl4

Contact [7]

Answer:

We assume you are converting between moles CCl4 and gram. You can view more details on each measurement unit: molecular weight of CCl4 or grams This compound is also known as Carbon Tetrachloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CCl4, or 153.8227 grams.

Explanation:

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3 0

2 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

(a) As per question, lead is oxidized and copper is reduced.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

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

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

(b) As per question, lead is reduced and copper is oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

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

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

Power gathered from the sun and wind

Mademuasel [1]

Answer: B. Can provide humans with energy forever

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3 years ago

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Lynna [10]

Answer:

I think It's C

Explanation:

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2 years ago

What determines an element's chemical properties

Alik [6]

The three factors determine the chemical properties of an element:
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2 years ago

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