Two atoms with an electronegativity difference of 0.4 form a bond that is
1. ionic, because electrons are shared
2. ionic, because electrons are transferred
3. covalent, because electrons are shared
4. covalent, because electrons are transferred

The correct answer is 3.

Explanation:

Ionic bond is defined as the bond formed when there is complete transfer of electrons from one atom to another atom. Example: NaCl

Covalent bond is defined as the bond formed when there is sharing of electrons between two atoms. The atom which takes the shared pair of electron towards itself is known as electronegative atom. Example: HCl

Dependency of electronegativity difference on ionic and covalent polar bond:

If the electronegativity difference is greater than 0 and less than 2.0, then the bond is covalent polar in nature and it shares the electrons.

And if the electronegativity difference is 2.0 or greater, then the bond is ionic polar in nature and the electrons are transferred from one atom to another.

As, the electronegativity difference given is 0.4, so the bond will be covalent and sharing of electrons will be there.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol