The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
Given question is incomplete. The complete question is as follows.
The successive ionization energies of a certain third-period element are I1 = 577.9KJ/mol, I2 + 1820 KJ/mol, I3 = 2750 KJ/mol, I4 = 11600 KJ/mol, I5 = 14800 KJ/mol. what element do these ionization energies suggest? Explain your reasoning.
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
Here, given ionization energies belong to an element present in third period.
We know that second ionization energy will always be greater than third ionization energy.
For the given ionization energies, there is a huge difference between third and fourth ionization energy. This means that there are three valence electrons present in the element.
This is because after losing three electrons it is difficult for the given element to lose fourth electron. Hence, is high as compared to .
Hence, this element has 3 valence electrons and it belongs to 3A group of the periodic table.
Thus, we can conclude that the given unknown element is aluminium (Al).
The table tells us that the larger the diameter the shorter the period of rotation is
The acetic acid will dissociate as
We will use Hendersen Hassalbalch's equation here
pH = pKa + log [salt] / [acid]
Where
[Acid] = concentration of protonated form
5.1 = 4.8 + log [salt] / [acid]
1.99= [salt] / [acid]
[salt] +[acid] = 0.5
1.99 [acid] + [acid] = 0.5
[acid] = 0.167 M
[acid] / total concentration = 0.167 / 0.5 = 0.33.4
So percentage = 33.4 % or approx 33 %
Answer:
l guess c is the answer
Explanation:
it is true because in oxygen does not share any atoms with hydrogen