Answer:
Element Lithium
Explanation:
The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals
It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.
Its atomic number is 3. The electronic configuration is 1s2 2S1
Explanation:
36.5gm (Molar mass) = 1mole
1 gram = 1/36.5 mole
10grams =(1/36.5) x 10 moles
= 0.027 x 10 moles
=0.27moles
If the half life is 5 days, you would have to divide 222 by 2 3 times.
so 222/2 =111g
111/2 = 55.5g
55.5/2 = 27.75g
after four half life the mass would be
27.75/2
13.875g
the percent would be
13.875/222
= 6.25%
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
