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>
A. Cuz it contains more salt
Hope this helps good luck on ur examsss :))))
Answer is: Cl and Na.
sodium and chlorine are in third period and they have very different properties. Sodium is solid metal and chlorine is gaseous nonmetal.
They form compound NaCl (Sodium chloride), because sodium lost one valence electron and form cation Na⁺, chlorine gain one electron and form anion Cl⁻.
Electron configuration of sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.
Electron configuration of chlorine atom: ₁₇Cl 1s² 2s² 2p⁶ 3s² 3p⁵.
Other examples are metal-metal pairs and they do not form cation and anion.
