Answer:
6.72M of HNO3
Explanation:
In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:
290.7mL + 350.0mL = 640.7mL
And you are diluting the solution:
640.7mL / 350.0mL = 1.8306 times
As the original concentration was 12.3M, the final concentration will be:
12.3M / 1.8306 =
<h3>6.72M of HNO3</h3>
Answer: D.Aluminium Oxide 0.10, Magnesium Oxide 0.50
Explanation:
Number of moles of NaOH= number of moles × volume
Number of moles= 100/1000 × 2 = 0.2 moles
Since;
2 moles of NaOH yield 1 mole of Al2O3
0.2 moles of NaOH will yield 0.2 × 1/2 = 0.1 moles of Al2O3.
Number of moles of HCl= 800/1000 × 2 = 1.6 moles
If 1 mole of Al2O3 requires 6 moles of HCl
0.1 moles of Al2O3 requires 0.1 × 6 = 0.6 moles of HCl.
Number of moles of HCl left after reaction with Al2O3 = 1.6- 0.6 = 1 mole
This leftover reacts with MgO
But;
1 mole of MgO reacts with 2 moles of HCl
x moles of MgO reacts with 1 mole of HCl
Thus; x= 0.5 moles of MgO
10 molecules of water are needed.
Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol