<span>Ethoxyethane; trifluoroborane; BF3.Et2O; Boron trifluoride ethyl ether; Boron trifluoride diethyl ether; Boron trifluoride-diethyl ether; Boron
</span>
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
Answer:
See Explanation
Explanation:
Ionization energy refers to the energy required to remove an electron from an atom. Metals have lower ionization energy than non metals since ionization energy increases across a period.
One thing that we must have in mind is that it takes much more energy to remove an electron from an inner filled shell than it takes to remove an electron from an outermost incompletely filled shell.
Now let us consider the case of magnesium which has two outermost electrons. Between IE2 and IE3 we have now moved to an inner filled shell(IE3 refers to removal of electrons from the inner second shell) and a lot of energy is required to remove an electron from this inner filled shell, hence the jump.
For aluminium having three outermost electrons, there is a jump between IE3 and IE4 because IE4 deals with electron removal from a second inner filled shell and a lot of energy is involved in the process hence the jump.
Hence a jump occurs each time electrons are removed from an inner filled shell.
