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.
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.
Molarity is defined as the number of moles present in one liter of solution. Given the mass of NH₃ is 2.35 g and its molar mass is 17 g/mol, the moles of NH₃ in 2.35 g can be calculated as 2.35 g / 17 g/mol = 0.138 mol. Consequently, in a 0.05 L solution, the number of moles amounts to 0.138 mol. Therefore, the concentration in 1 L is: 0.138 mol / 0.05 L x 1L = 2.76 mol. Thus, the molarity of NH₃ is 2.76 M.
The amount of oxygen atoms present is approximately 3.27·10²³. To determine this figure, we must first assess the sodium sulfate sample. The chemical formula for it is Na₂SO₄, which possesses a molar mass of roughly 142.05 g/mol. We can then use stoichiometry to convert the mass of Na₂SO₄ into moles. By knowing the moles of Na₂SO₄, we will subsequently convert this to moles of oxygen utilizing the mole ratio and finally apply Avogadro's number to convert to atoms of oxygen. Thus, with the calculations completed, the resulting quantity of oxygen atoms is about 3.27·10²³.
