By stoichiometry and assume
that:
CxH2xOy + zO2 -> xCO2
+ xH2O
<span>
CO2: 9.48/44 = 0.215 mmol
H2O: 3.87/18 = 0.215 mmol
mass of C = 0.215 * 12 = 2.58 mg
mass of H = 0.215 * 2 * 1 = 0.43 mg
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg
So C/O = 2.58/1.11 ≈ 3 </span>
<span>
Thus we have C3H6O</span>
<span> </span>
Answer:
41.67 mol
Explanation:
1 Litre of water = 1000g
Mole = mass / molar mass
Mass of 1 L of water = 1000 g
Molar mass of water (H2O) :
(H = 1, O = 16)
H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol
Amount of water consumed = (3/4) of 1 litre
= (3/4) * 1000g
= 750g
Therefore mass of water consumed = 750g
Mole = 750g / 18g/mol
Mole of water consumed = 41.6666
= 41.67 mol
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.
