Yes what the other person said can I plz get an thanks
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
In comparison see it is very easy in goolge
1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
Learn more about empirical formula:
brainly.com/question/24297883
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<span>(2.00 L) x (0.100 M HCl) / (1.50 M HCl) = 0.133 L</span>
