Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
Answer:
5.0 38 84.0 749.7 528.0 729.0 738.9 739.0
Ewdrqw3ecdr4qwe4rfwqefq343rfqweftrweqf
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
