Answer:
molar mass = 180.833 g/mol
Explanation:
- mass sln = mass solute + mass solvent
∴ solute: unknown molecular (nonelectrolyte)
∴ solvent: water
∴ mass solute = 17.5 g
∴ mass solvent = 100.0 g = 0.1 Kg
⇒ mass sln = 117.5 g
freezing point:
∴ ΔTc = -1.8 °C
∴ Kc H2O = 1.86 °C.Kg/mol
∴ m: molality (mol solute/Kg solvent)
⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)
⇒ m = 0.9677 mol solute/Kg solvent
- molar mass (Mw) [=] g/mol
∴ mol solute = ( m )×(Kg solvent)
⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )
⇒ mol solute = 0.09677 mol
⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )
⇒ Mw solute = 180.833 g/mol
Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
Answer:
73.46839716589713698731965
Explanation:
Answer:
Ruler
Explanation:
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