Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=
Answer:
1.772 gram is the approximate answer
Explanation:
molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole
the reaction is
AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl
from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3
implies 132g AlCl3 gives 78g Al(OH)3
Implies 3g AlCl3 gives
3*122/78 = 1.772 grams