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vekshin1
3 years ago
6

If two atoms have an electronegativity difference of 1.3, what type of bond would form between them?

Chemistry
2 answers:
Debora [2.8K]3 years ago
5 0

Answer:

The correct answer is b polar covalent

Explanation:

When two atoms joined by covalent bond has difference in their electronegativities at that time polarity arise.

    When the electronegativity difference is low such as 1.3 then the polar bond formed by the two atoms are called polar covalent bond.For example H2O

 on the other hand polar bond formed by two atoms having high difference in their electronegativities is called ionic bond.For example NaCl.

zloy xaker [14]3 years ago
3 0

Answer:

a. polar covalent

Explanation:

Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.

The higher the electronegativity, the more an atom attracts electrons towards itself.

The most commonly used method of calculation of electronegativity is done by Pauling's scale.

Cesium is the least electronegative (0.79), and fluorine is most electronegative (3.98)

Atoms that form bonds with high electronegativity difference among them form ionic bonds (metal and non-metal). If their electronegativity difference (Δ)=0, then they form non polar covalent bond.

  • Δ=0 : Non- polar covalent bond.
  • 0 < Δ < 1.7 : Polar covalent bond.
  • Δ = 1.7 : 50% ionic and 50% covalent bond
  • Δ > 1.7 : Ionic bond

The answer is a. polar covalent

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The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

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}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
What mass of K2CO3 is needed to prepare 200. mL of a solution having a potassium ion concentration of 0.150 M
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<u>Answer:</u>

2.07 grams

<u>Explanation:</u>

We know that the molecular mass of K_2CO_3 is 138.205 grams.

In order to find out how many grams we need to make 200 mL having a potassium ion concentration of 0.150, we need to find the number of moles first.

M = moles / liters

0.150 = x / 0.200

x = 0.03 moles

Since there are 2 potassium atoms in each molecule of K_2CO_3 so we will divide this number by half and multiply it by the molecular mass.

0.03/2 = 0.015 moles

Mass of K_2CO_3 needed = 0.015 × 138.205 = 2.07 grams

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