Answer: The correct selection is (b).
Explanation:
The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.
This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.
Therefore, a significant amount of energy is needed to dislodge the valence electrons.
The electronic configuration for helium is 
. Hence, due to its fully occupied valence shell, it exhibits greater stability.
Consequently, a large amount of energy is needed to remove an electron from a helium atom.
In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.
The answer is true. A solid solution consists of a solid state solution formed by one or more solutes dissolved in a solvent, or a combination of two crystalline solids that coexist within a crystal lattice. Metal alloys, semiconductors, and moist solids are examples of such solid solutions.
We need to calculate the volume of Gold, assuming its mass matches that of copper.
Given information:
Density of Copper = 8.96 g/ml.
Volume of Copper = 141 ml.
Mass of Gold = Mass of Copper.
Density of Gold = 19.3 g/ml.
To find copper's mass, we use the density equation:
Density = mass/volume.
To find mass of copper:
Mass of copper = Density of Copper * Volume of Copper.
Mass of copper = 8.96 g/ml * 141 ml = 1263.36 g.
Thus,
Mass of gold = Mass of copper = 1263.36 g.
Now, using the density formula for gold to get its volume:
Volume of gold = Mass of gold / Density of gold.
Volume of gold = 1263.36 g / 19.3 g/ml = 65.46 mL.
Consequently, the volume of gold required to match the mass of copper is 65.46 mL.
The composition consists of 62 % one isomer and 38 % its enantiomer.
Assuming that the mixture comprises 62 % of the (R)-isomer.
Then the percentage of the (S) is calculated as 100 % - 62 % = 38 %.
Enantiomeric excess = % (R) - % (S) = 62 % - 38 % = 24 %.
The ore contains a 55.4% composition of calcium phosphate (related to apatite), leading to a calculation where Ca3(PO4)2 equals 55.4%x=1000g, resulting in x=1000/0.554, which equals 1.805kg. To find the percentage of phosphorus in this amount of calcium phosphate, calculate the total masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (noting that oxygen contributes 16 mass x 4 =64), giving a cumulative mass of 310.2, while the phosphorus is 61.95 (Pmass x 2). Therefore, 61.95/310.2= 0.19 or 19% for phosphorus. Consequently, from 1.805 x 0.19, there will be 0.34kg of phosphorus.
