Answer:
Aluminum
Explanation:
Given
<em>See attachment for chart</em>
Required
Identify the unknown substance
To do this, we simply calculate the specific heat capacity from the given parameters using:
This gives:
So, we have:
From the attached chart, we have:
--- The specific heat capacity of Aluminum
<em>Hence, the unknown substance is Aluminum</em>
Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
Answer:
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32 60
CH₃OH + CO ⟶ CH₃COOH
m/g: 160
(a) Moles of CH₃OH
(b) Moles of CH₃COOH
(c) Mass of CH₃COOH
Answer;
K+ and NO3- ions
Explanation;
The main ions remaining are K+ and NO3- ions after pbi2 precipitation is complete.
However; There will always be tiny amounts of Pb2+ and I- ions, but most of them are in the solid precipitate.
Answer:
Near the boiling point of the solvent
Explanation:
The process of recrystallization is hinged on the fact that the amount of solute that can be dissolved by a solvent increases with temperature. The process involves creation of a solution by dissolving a solute in a solvent at or near its boiling point. At the boiling point of the solvent, the solute has a greater solubility in the solvent; not much volume of the hot solvent is required to dissolve the solute.
Before the solution is later cooled, you can now filter out insoluble impurities from the hot solvent. The quantity of the original solute drops appreciably because impurities have been removed. At this lower temperature, the solution becomes saturated and the solute can no longer be held in solution hence it forms pure crystals of solute, which can be recovered.
Recrystallization must be carried out using the proper solvent. The solute must be relatively insoluble in the solvent at room temperature but more soluble in the solvent at elevated temperature.
