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.
Molar mass of LiBr (mm )= 86.845 g/mol
Molarity ( M ) = 4 M
Mass of solute ( m ) = 100 g
Volume ( V ) = in liters ?
V = m / mm * M
V = 100 / 86.845 * 4
V = 100 / 347.38
V = 0.2875 L
hope this helps!.
Answer:
53.7 grams of HNO3 will be produced
Explanation:
Step 1: Data given
Mass of NO2 = 59.0 grams
Molar mass NO2 = 46.0 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 59.0 grams / 46.0 g/mol
Moles NO2 = 1.28 moles
Step 4: Calculate moles HNO3
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3
Step 7: Calculate mass HNO3
Mass HNO3 = 0.853 moles * 63.01 g/mol
Mass HNO3 = 53.7 grams
53.7 grams of HNO3 will be produced
Answer:
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Explanation:
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>