Answer:
The answer to your question is letter D
Explanation:
Data
mass of Li = 4.5 g
mass of N₂ = 4.5 g
limiting reactant = ?
Reaction
6Li (s) + N₂ (g) ⇒ 2Li₃N
This reaction is balanced
Process
1.- Calculate the molecular mass of the reactants
Li = 7 x 6 = 42g
N = 14 x 2 = 28 g
Calculate the limiting reactant using proportions
Theoretical proportion Li / N = 42 / 28 = 1.5
Experimental proportion Li/N = 4.5/4.5 = 1
As the experimental proportion was lower than the theoretical, the limiting reactant is Lithium
2.- Calculate the theoretical yield of Li₃N
42 g of lithium -------------- 70 g of Li₃N
4.5 g of lithium ------------- x
x = (4.5 x 70) / 42
x = 315 / 42
<u> x = 7.5 %</u>
<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
The question is accompanied of a table showing the atomic masses and abundances of the five isotopes and asks to calculate the atomic mass of the element with five significant figures. You just have to add the products of the atomic mass of each isotope times its abundance and divide by 100. This is the calculation:<span>
(179.946706*0.12 + 181.948206*26.5 + 182.9502245*14.31 + 183.9509326*30.64 + 185.954362*28.43 ) / 100 = 183.84 (which is rounded to five significant figures)</span>
The mass number is protons and neutrons. So I think it’s false
The result is the sound will be louder. This is due to the high amount of energy in the sounds.
