Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer: all other conditions equal, the rate evaporation of a contained liquid will be slower than the rate of evaporation of an uncontained liquid.
Justification:
1) The rate of evaporation increases as the surface area of the liquid (relative to the whole content) increases. This is, the greater the surface is the faster the evaporation.
2) That is so because the higher the surface of the liquid the more the number of particles in the liquid that are in contact with the surrounding air and so the more the particles will escape from the liquid to the air (which is what evaporation is).
3) A liquid contained will take the form of the container, so part of the liquid wil remain below the surface, while an uncontained liquid will spread all over the surface and so pratically all the liquid is in contact witht the air surrounding it.
Answer:
The answer to your question is Volume = 0.12 L
Explanation:
- Data
Pressure 1 = 1 atm
Volume 1 = 1.82 L
Pressure 2 = 15 atm
Volume 2 = ?
Temperature = constant
- Formula
- To solve this problem use the Boyle's law
P1V1 = P2V2
- Solve for V2
V2 = P1V1 / P2
- Substitution
V2 = (1)(1.82) / 15
- Simplification
V2 = 1.82 / 15
- Result
V2 = 0.12 L