Assign oxidation numbers to the element in each of the following compounds. Remember an oxidation number is for one atom of the element. a. HgO
b. Al4C3
c. CrF2
d. Fe2S3

a.  x = + 2

b.  x = + 3

c.  x = + 2

d.  x = + 3

Explanation:

The oxidation number is a formal charge assigned to an atom present in a molecule or formula unit or ion based on some arbitrary rules.

a.

HgO

The oxidation number of Hg in HgO is:

x + (-2) = 0

x = +2

b.

Al₄C₃

The oxidation number of Al in Al₄C₃ is:

4x +(3 × -4) = 0

4x - 12 = 0

4x = +12

x = 12/4

x = +3

c. CrF₂

x + ( 2 × - 1) = 0

x - 2 = 0

x = + 2

d. Fe₂S₃

2x + ( 3 × - 2) = 0

2x + (-6) = 0

2x = 6

x = 6/2

x = +3

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j