Which element would most likely have chemical properties similar to that of fluorine (f)? ne n li br

The correct answer is Bromine.

Explanation:

Elements are distributed into 7 Periods and 18 Groups. The elements which have same number valence electrons or which belong to same group will show similar chemical properties.

Fluorine is the element which belongs to Group 17. The electronic configuration of this element is The number of valence electrons of this element is 7.

For the given options:

Option 1: Ne

This element belongs to Group 18. The electronic configuration is: The number of valence electrons of this element is 8.

Option 2: N

This element belongs to Group 15. The electronic configuration is: The number of valence electrons of this element is 5.

Option 3: Li

This element belongs to Group 1. The electronic configuration is: The number of valence electrons of this element is 1.

Option 4: Br

This element belongs to Group 18. The electronic configuration is: The number of valence electrons of this element is 7.

From the above information, we can conclude that the element which shows similar chemical properties to fluorine is Bromine.

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