Answer:
Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.78 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Explanation:
The mass of nonvolatile solute added is ---- 29.2g
The mass of solvent benzene is ---- 0.250kg = 250g
The Kf value of benzene is ---- 5.12^oC/m.
Depression in the freezing point of the solution is --- 2.78^oC.
What is the molar mass of the unknown solute?
Substitute the given values in this formula to get the molar mass of unknown solvent:
Hence, the molar mass of unknown solute is --- 215g/mol.
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Bellow you my find the chemical equations of the ionization of a triprotic acid.
Explanation:
The chemical equations for the ionization of phosphoric acid (H₃PO₄), which is a triprotic acid, in water:
H₃PO₄ (s) + H₂O (l) → H₂PO₄⁻ (aq) + H₃O⁺ (aq)
H₂PO₄⁻ (aq) + H₂O (l) → HPO₄²⁻ (aq) + H₃O⁺ (aq)
HPO₄²⁻ (aq) + H₂O (l) → PO₄³⁻ (aq) + H₃O⁺ (aq)
where:
(s) - solid
(l) - liquid
(aq) - aqueous
Learn more about:
triprotic acid
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