Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
Moles of Zn: 26 / 65 = 0.4
Moles of S: 12.8 / 32 = 0.4
Molar ratio of Zn : S = 1 : 1
Empircal formula: ZnS
The answer is C
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
Answer:
665 g
Explanation:
Let's consider the following thermochemical equation.
2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol
According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:
-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂
The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:
15.1 mol × 44.01 g/mol = 665 g
Can you include the pome pls
