Answer:
470 °C
Explanation:
This looks like a case where we can use Charles’ Law:
Data:
V₁ = 20 L; T₁ = 100 °C
V₂ = 40 L; T₂ = ?
Calculations:
(a) Convert the temperature to kelvins
T₁ = (100 + 273.15) K = 373.15 K
(b) Calculate the new temperature
Note: The answer can have only two significant figures because that is all you gave for the volumes.
(c) Convert the temperature to Celsius
T₂ = (750 – 273.15) °C = 470 °C
Answer:
8.66 g of Al₂O₃ will be produced
Explanation:
4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
This is the reaction.
Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.
4.6 g / 26.98 g/mol = 0.170 moles
Ratio is 4:2.
4 moles of aluminum can produce 2 moles of Al₂O₃
0.170 moles of Al, may produce (0.170 .2)/ 4 = 0.085 moles
Let's convert the moles of Al₂O₃ to mass.
0.085 mol . 101.96 g/mol = 8.66 g
You can boil the salt water. The water will evaporate while the salt will be left behind because it's a solid. Salt has a much higher boiling point than water (2,575° F), so that's why it won't evaporate with the water.
