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
Answer:
D
Explanation: A precipitation reaction is a type of chemical reaction in which two soluble salts in aqueous solution combine and one of the products is an insoluble salt called a precipitate.
A map scale indicates the size of a mapped area.
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
Answer:
A
Explanation:
increase in energy - hotter and/or more reactive
decrease in energy - colder and/or less reactive
Since it is going from a liquid to solid it will have a tighter arrangement of particles.