Answer:Radium
Explanation:
The nuclear reaction involving two alpha emissions of 234 U is shown in the diagram. This leads to the formation of a 226Ra nucleus.
Answer:
Graham's law states that the rate of effusion or of diffusion of a gas is inversely proportional to the square root of its molecular weight. So, rate of diffusion of hydrogen gas is four times faster than oxygen gas.
Explanation:
Answer:
(a) 1s2 2s1
Explanation:
Electron configurations of atoms are in their ground state when the electrons completely fill each orbital before starting to fill the next orbital.
<h3><u>
Understanding the notation</u></h3>
It's important to know how to read and interpret the notation.
For example, the first part of option (a) says "1s2"
- The "1" means the first level or shell
- The "s" means in an s-orbital
- The "2" means there are 2 electrons in that orbital
<h3><u>
</u></h3><h3><u>
Other things to know about electron orbitals</u></h3>
It important to know which orbitals are in each shell:
- In level 1, there is only an s-orbital
- In level 2, there is an s-orbital and a p-orbital
- in level 3, there is an s-orbital, a p-orbital, and a d-orbital <em>(things get a little tricky when the d-orbitals get involved, but this problem is checking on the basic concept -- not the higher level trickery)</em>
So, it's also important to know how many electrons can be in each orbital in order to know if they are full or not. The electrons should fill up these orbitals for each level, in this order:
- s-orbitals can hold 2
- p-orbitals can hold 6
- d-orbitals can hold 10 <em>(but again, that's beyond the scope of this problem)</em>
<h3><u>
Examining how the electrons are filling the orbitals</u></h3>
<u>For option (a):</u>
- the 1s orbital is filled with 2, and
- the 2s orbital has a single electron in it with no other orbitals involved.
This is in it's ground state.
<u>For option (b):</u>
- the 1s orbital is filled with 2,
- the 2s orbital is filled with 2,
- the 2p orbital has 5 (short of a full 6), and
- the 3s orbital has a single electron in it.
Because the 3s orbital has an electron, but the lower 2p before it isn't full. This is NOT in it's ground state.
<u>For option (c):</u>
- the 1s orbital is filled with 2,
- the 2s orbital has 1 (short of a full 2), and
- the 2p orbital is filled with 6
Although the 2p orbital is full, since the 2s orbital before it was not yet full, this is NOT in it's ground state.
<u>For option (d):</u>
- the 1s orbital has 1 (short of a full 2), and
- the 2s orbital is filled with 2
Again, despite that the final orbital (in this case, the 2s orbital), is full, since the 1s orbital before it was not yet full, this is NOT in it's ground state.
Given data:
Sublimation of K
K(s) ↔ K(g) ΔH(sub) = 89.0 kj/mol
Ionization energy for K
K(s) → K⁺ + e⁻ IE(K) = 419 Kj/mol
Electron affinity for Cl
Cl(g) + e⁻ → Cl⁻ EA(Cl) = -349 kj/mol
Bond energy for Cl₂
1/2Cl₂ (g) → Cl Bond energy = 243/2 = 121.5 kj/mol
Formation of KCl
K(s) + 1/2Cl₂(g) → KCl(s) ΔHf = -436.5 kJ/mol
<u>To determine:</u>
Lattice energy of KCl
K⁺(g) + Cl⁻(g) → KCl (s) U(KCl) = ?
<u>Explanation:</u>
The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i.e.
ΔHf(KCl) = U(KCl) + ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)
therefore:
U(KCl) = ΔHf(KCl) - [ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)]
= -436.5 - [89 + 419 + 243/2 -349] = -717 kJ/mol
Ans: the lattice energy of KCl = -717 kj/mol
