The boiling point and distillation temperature of a substance are the same. The answer would be True
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
The element is Sodium with an atomic number of 11 and electrovalent bonding takes place when it comes near an atom having seven valence electrons.
<h3>What is Electrovalent bonding?</h3>
This is also referred to as ionic bonding and involves the transfer of atoms of an element to another.
In order for both of them to achieve a stable octet configuration, sodium donates one atom to the element seven valence electrons.
Read more about Electrovalent bonding here brainly.com/question/1979431
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The value of Kc for the thermal decomposition of H₂S is 2.2 x 10⁻⁴ at 1400 K:
2 H₂S(g) ↔ 2 H₂(g) + S₂(g)
initial 3.5 M 0 0
at equilibrium 3.5 M - 2x 2x x
Kc = [S₂][H₂]² / [H₂S]²
2.2 X 10⁻⁴ = x(2x)² / (3.5 - 2x)²
2.2 x 10⁻⁴ = 4 x³ / (3.5)² Assuming x <<<<< 3.5
x = 0.088
Thus [H₂S] = 3.324 M
