Answer:
it went under the process known as crystalization
Explanation:
am a chemist
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
Answer:
CoCl₃ > Li₂SO₄ > NH₄I.
Explanation:
- Adding solute to water causes depression of the boiling point.
- The elevation in boiling point (ΔTf) can be calculated using the relation:
<em>ΔTf = i.Kf.m,</em>
where, ΔTf is the depression in freezing point.
i is the van 't Hoff factor.
- <em>van 't Hoff factor</em><em> is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.</em>
Kf is the molal depression constant of water.
m is the molality of the solution.
<u><em>(1) Li₂SO₄:</em></u>
i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.
∴ ΔTb for (Li₂SO₄) = i.Kb.m = (3)(Kf)(m) = 3(Kf)(m).
<u><em>(2) NH₄I:</em></u>
i for NH₄I = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.
∴ ΔTb for (NH₄I) = i.Kb.m = (2)(Kf)(m) = 2(Kf)(m).
<u><em>(3) CoCl₃:</em></u>
i for CoCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.
∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(m) = 4(Kf)(m).
- <em>So, the ranking of the freezing point from the highest to the lowest is:</em>
CoCl₃ > Li₂SO₄ > NH₄I.
