Apply the given electronegativities to each bond in the examples below to see which are polar and which are nonpolar. multiple bonds are treated the same as single bonds.
h-o-h
o-c-o
h-c-n

H-O-H  polar

O-C-O nonpolar

H-C-N   polar

Explanation:

Looking up the electronegativities of the atoms involved in this question, we have:

Atom   Electronegativity

H               2.2

C               2.55

N               3.04

O               3.44

All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.

In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H  of the ideal tetrahedral structure to be smaller than 109.5 º  resulting in an overall dipole moment making it polar.

In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule  since the central carbon atom does not have lone pairs of electrons since it  has 2 double bonds. Therefore the molecule is nonpolar.

In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.

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