<u>Given:</u>
Surface area at the narrow end, A1 = 5.00 cm2
Force applied at the narrow end, F1 = 81.0 N
Surface area at the wide end, A2 = 725 cm2
<u>To determine:</u>
Force F2 applied at the wide end
<u>Explanation:</u>
Use the relation
F1/A1 = F2/A2
F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N
Ans: (b)
The force applied at the wide end = 11,745 N
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
Hydrogen, Nitrogen, Flourine, Oxygen, Iodine, Chlorine, Iodine, and Bromine.
Answer:
A chemical reaction that has a positive ΔH is said to be endothermic , while a chemical reaction that has a negative ΔH is said to be exothermic . What does it mean if the ΔH of a process is positive? It means that the system in which the chemical reaction is occurring is gaining energy.
