-2 for O
-1 for F
0 for N2
+1 for Li
+3 for Fe
Answer:
The molar mas of the X is 203.06 g/mol.
Explanation:
The pressure of the pure solvent = p = 23.8 Torr
Vapor pressure of the solution =
Mass of solute =14.4 g
Molar mass of solute = M
Moles of solute =
Mass of solvent or water = 100.0 g
Moles of water =
Mole fraction of solute =
The relative lowering in vapor pressure of the solution with non volatile solute is equal mole fraction of solute in solution.
Solving for
The molar mas of the X is 203.06 g/mol.
Answer:
yh
Explanation:
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<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO