Answer:
89.88 g
Explanation:
Atomic Mass of Ar: 39.948
Mass = moles * AM
Replacing moles = 2.25 and AM = 39.948 you get the mass of Ar:
Mass = 2.25 * 39.948
Mass = 89.88 g
Answer:
H₂: 0.48, N₂: 0.43; Ar: 0.09
Explanation:
First of all, sum all the pressures to know the total pressure in the mixture.
434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr
Mole fraction = Pressure gas / Total Pressure
Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48
Mole Fraction N₂: 389.9 /901.8 Torr =0.43
Mole Fraction Ar: 77.9 /901.8 Torr = 0.09
Remember: <u>SUM OF MOLE FRACTION = 1</u>
Answer:
The answer to your question is 1.11 M
Explanation:
Data
volume 1 = 287 ml
concentration 1 = 1.6 M
volume 2= 412 ml
concentration 2 = ?
Formula
Volume 1 x concentration 1 = Volume 2 x concentration 2
Solve for concentration 2
concentration 2 = (volume 1 x concentration 1) / volume 2
Substitution
concentration 2 = (287 x 1.6) / 412
Simplification
concentration 2 = 459.2 / 412
Result
concentration 2 = 1.11 M
Answer:107.1 g, 124.1 g
Explanation:
The equation of the reaction is;
Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)
Hence;
For Al2S3
Number of moles= reacting mass/molar mass
Number of moles = 158g/150gmol-1 =1.05 moles
If 1 mole of Al2S3 yields 3 moles of H2S
1.05 moles of Al2S will yield
1.05 × 3/1 = 3.15 moles
Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g
For water
Number of moles of water = 131g/18gmol-1= 7.3 moles
6 moles of water yields 3 moles of H2S
7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S
3.65 moles × 34 gmol-1 =124.1 g
