This answer is based on the electron configuration.
And you can use Aufbau's rule to predict the atomic number of the next elements.
Radon, Rn is the element number 86.
Following Aufbau's rules, the electron configuration of Rn is: [Xe] 6s2 4f14 5d10 6p6. This means that you are suming 2 + 14 + 10 + 6 = 32 electrons with respect to the element Xe.
You can verity that the atomic number of Xe is 54, so when you add 32 you get 54 + 32 = 86, which is the atomic number of Rn.
Again, as per Aufbau's rules, the next element of the same group or period is when the 6 electrons of the 7p orbital are filled. For that, they have to pass 32 elements whose orbitals are:
7s2 5f14 6d10 7p6: count the electrons added: 2 + 14 + 10 + 6 = 32, and that is why the next element wil have atomic number 86 + 32 = 118.
Now, when you go for a new series, you find a new type of orbital, the g orbital, for which the model predict there are 18 electrons to fill.
So the next element of the group will have ; 2 + 18 + 14 + 10 + 6 = 50 electrons, which means that the atomic number of this, not yet discovered element, has atomic number 118 + 50 = 168.
By the way the element with atomic number 118 was already discovdered at its symbol is Og. You can search that information in internet.
Answers: 118 and 168
Answer:
<h3>The answer is 320.75 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>320.75 mL</h3>
Hope this helps you
Answer:
The answer to your question is: ΔHrxm = -23.9 kJ
Explanation:
Data:
2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ (1)
CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ (2)
Reaction:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
We invert (1) and change the sign of ΔH
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
We multiply (2) by 3
3( CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ) (2)
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
We add (1) and (2)
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
Fe2O3(s) + 3CO(g)+3/2O2(g) → 2Fe(s)+3/2O2 + 3CO2(g)
Simplify
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) and ΔHrxm = -23.9 kJ
The answer is A as they had to spend time growing or looking for food as food was scarce back then.
