Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

For this, it is necessary to know the values in meters for any of these diameters:


Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.

<h3>Diameter of a biscuit in meters</h3>

<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:





In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
By using the ICE table :
initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X
when Ka = (X) (X) / (0.2-X)
so by substitution:
4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6
and PH = -㏒[H+]
= -㏒ 10^-6
= 6
Answer: A) Sodium Na and Magnesium Mg
Explanation:
In the period table, you can see that Sodium (group 1) has 11 protons, and Magnesium (group 2) has 12 protons. 11+12=23 protons
Answer:
Oxygen is in group 16/VIA, which is called the chalcogens, and members of the same group have similar properties. Sulfur and selenium are the next two elements in the group, and they react with hydrogen gas (H2) in a manner similar to oxygen.
Explanation: