The visual representation is displayed in the following image.
For calculations, consider 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; this represents the mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; indicating the mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; this is the quantity of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; this is the quantity of carbon.
n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.
The compound in question is identified as dichlorocarbene CCl₂.
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.
Answer: The mole fraction of hydrogen gas at 20°C is 0.975
Explanation:
The information provided includes:
Water vapor pressure at 20°C is 17.5 torr
Total pressure at 20°C = 700.0 torr
Hydrogen gas vapor pressure at 20°C = (700.0 - 17.5) torr = 682.5 torr
To find hydrogen gas's mole fraction at 20°C, we utilize Raoult's law, represented by:
where,
= pressure of hydrogen gas = 682.5 torr
= total pressure = 700.0 torr
= mole fraction of hydrogen gas =?
Substituting the values into the equation yields:
Thus, the mole fraction of hydrogen gas at 20°C equals 0.975
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
