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:
sediment can only form by weathering, erosion, and deposition
Answer:
C. Alkaline Metal ok thanks
Answer:
T₂ = 221.8 K
Explanation:
Given data:
Initial temperature = 57.9°C ( 57.9+273 =330.9 k)
Initial volume = 75.8 mL
Final volume = 50.8 mL
Final temperature = ?
Solution:
According to Charles's law,
V₁ /T₁= V₂/T₂
T₂ = V₂T₁/V₁
T₂ = 50.8 mL ×330.9 k / 75.8 mL
T₂ = 16809.72 mL.K/ 75.8 mL
T₂ = 221.8 K
Answer:
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
Explanation:
molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.
Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .
So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .
Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .
