All categories

3 years ago

When an ionic bond forms between lithium (Li) and fluorine (F), which of the following occurs?

Chemistry

2 answers:

Zigmanuir [339]3 years ago

6 0

Answer:

F gains one electron.

Explanation:

Hello,

Due to the high electronegativity of the fluorine atom (approx 3.98), it tends to gain electrons rather than lose them. Now, as lithium has the following electron configuration:

$1s^22s^1$

It loses one electron that is gained by the fluorine to form lithium fluoride, LiF.

Best regards.

allochka39001 [22]3 years ago

3 0

They don't share and both deal with only 1 electron.

Lithium gives away 1 electron.
F will receive that electron.

The chemical formula is LiF

You might be interested in

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

How much energy is required to heat 2kg of ice from -5°c to 0°c

balu736 [363]

Explanation:

total heat = Heat required to convert 2 kg of ice to 2 kg of water at 0 °C + Heat required to convert 2 kg of water at 0 °C to 2 kg of water at 20 °C.

Heat=mhfg+mCpΔT

Here, m ( mass of ice) = 2 kg

hfg (latent heat of fusion of ice) = 334 KJ

Cp of water (specific heat) = 4.187 KJ/Kg-K

ΔT(Temperature difference) = 20 °C

Therefore, Heat required = 2 x 334 + 2 x 4.187 x (20 - 0 )

Heat reqd= 835.48 KJ

Therefore, to melt 2 kg of ice 835.48 KJ of heat is required.

4 0

2 years ago

Select which of the following are used to determine the age of the Earth. Select 3.

saveliy_v [14]

Meteorites discovered on earth

7 0

2 years ago

A sample of PCl5(g) was placed in an otherwise empty flask at an initial pressure of 0.500 atm and a temperature above 500 K. Ov

tia_tia [17]

Answer:

Kp = 0.81666

Explanation:

Pressure of PCl₅ = 0.500 atm

Considering the ICE table for the equilibrium as:

PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)

t = o 0.500

t = eq -x x x

--------------------------------------------- --------------------------

Moles at eq: 0.500-x x x

Given the pressure of PCl₅ at equilibrium = 0.150 atm

Thus, 0.500 - x = 0.150

x = 0.350 atm

The expression for the equilibrium constant is:

$K_p=\frac {P_{PCl_3}P_{[Cl_2}}{P_{PCl_5}}$

So,

$K_p=\frac{x^2}{0.500-x}$

x = 0.350 atm

Thus,

$K_p=\frac{{0.350}^2}{0.500-0.350}$

Thus, Kp = 0.81666

7 0

3 years ago

Indicate which of the following statements is FALSE.a. Covalent bonds connect nucleotides in a strand; noncovalent interactions

LenaWriter [7]

Answer:

The false statement is d Avery,Macleod and McCarty showed that DNA is the genetic information of cells and RNA is the genetic information in the viruses .

Explanation:

Avery,Macleod and MacCarty showed that DNA is the genetic material of the cell.

On the other hand Fraenkel, Conrat and Sanger carried out their experiment on tobacco mosaic virus to prove that RNA act as genetic material in some viruses.

8 0

3 years ago