Answer:
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
Because it has 4 electrons in outermost shell.
There are 18 numbered groups in the periodic table, and the f-block columns (between groups 3 and 4) are not numbered. The elements in a group have similar physical or chemical characteristics of the outermost electron shells of their atoms (i.e., the same core charge), as most chemical properties are dominated by the orbital location of the outermost electron. There are three systems of group numbering. The modern numbering group 1 to group 18 is recommended by the International Union of Pure and Applied Chemistry (IUPAC). It replaces two older naming schemes that were mutually confusing. Also, groups may be identified by their topmost element or have a specific name. For example, group 16 is variously described as the oxygen group and as the chalcogens.
Answer:
1.56 g of water was involved in the reaction
Explanation:
From the stoichiometric equation
2Na + 2H2O = 2NaOH + H2
NB : Mm Na= 23, Mm H2O = ( 2+16)= 18
2(23) of Na requires 2(18) of water
Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water
...a satellite!!!! that is one and maybe only answer
