Use the standard reduction potentials located in the Tables' linked above to calculate the equilibrium constant for the reaction: Cd^2+(aq) + H2(g) >Cd(s) + 2H+(aq)

3.1 * 10^-14

Explanation:

Note that E°cell = 0.0592/n log K

We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen

Anode reaction

H2(g) > 2H+ + 2e

Cathode reaction

Cd^2+(aq) + 2e > Cd(s)

E°cell = E°cathode - E°anode

E°cathode = –0.40 V

E°anode = 0 V

E°cell =  –0.40 V - 0 V

E°cell = –0.40 V

E°cell = 0.0592/n log K

Where n=2 electrons transferred

–0.40 = 0.0592/2 log K

–0.40 = 0.0296 log K

log K = –0.40/0.0296

log K = -13.5135

K = Antilog ( -13.5135)

K = 3.1 * 10^-14

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied