2) mg donates two protons to O.
Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
Independent Variable: a variable that you can change in an experiment
Dependent Variable: something that changes as you change the independent variable
control variable: something that is not changed throughout the experiment
Answer:
Potassium selenide
Explanation:
Potassium selenide (K2Se)
The answer is 4.9 moles.
Solution:
Using the equation for boiling point elevation Δt,
Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,
m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
m = number of moles / 1.0kg
number of moles = m*1.0kg
= (4.883 mol/kg) * (1.0kg)
= 4.9 moles
