Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

Explanation:

It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.

              or

Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.

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.

Refer to the explanation and the attached image for further details. When AlCl3 reacts with (CH3)3CCH2Cl, a primary carbocation is produced. This primary carbocation then undergoes a 1,2-alkyl shift leading to the formation of a tertiary carbocation, which subsequently bonds with the benzene ring as depicted in the attached image.