What are the trends and exceptions to the trends in electron affinity? check all that apply. the electron affinities of the elements in group 17 are larger (more negative) than the elements in group 1. electron affinities increase (become more negative) down a group. elements in group 14 have larger (more negative) electron affinities than elements in group 15. elements in group 2 have very large electron affinities.

the correct answers are:

1) The electron affinities of the elements in Group 17 are larger (more negative) than the elements in Group 1.

3) Elements in Group 14 have larger (more negative) electron affinities than elements in Group 15.

Explanation:

When an electron is added to a neutral atom to convert it into a negative ion then the amount of change taking place in its energy is known as electron affinity.

Smaller is the size of an atom more readily an electron will be gained by it because of strong force of attraction exerted by its nucleus towards the electron.

So, on moving down a group there will occur a decrease in electron affinity of the atoms. Also, size of atoms decreases from left to right in a periodic table. Therefore, electron affinity will also increase on moving from left to right.

As elements of group 15 are half filled so, they are more stable in nature and have slightly less electron affinity as compared to group 14 elements.

Thus, we can conclude that the trends and exceptions to the trends in electron affinity are as follows.

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