Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Workplace Hazardous Materials Information System is the answer to this question. Hope it helps :)
Answer:
The fluoride which precipitates first is CaF₂
Explanation:
When F⁻ is added, CaF₂ and BaF₂ are produced following the ksp equation:
For CaF₂:
Ksp = 3.2x10⁻¹¹ = [Ca²⁺] [F⁻]²
<em>Where [Ca²⁺] = 0.075M * {35mL / (25mL + 35mL)} = 0.04375M</em>
3.2x10⁻¹¹ = [0.04375M] [F⁻]²
[F⁻]² = 7.31x10⁻¹⁰
[F⁻] = 2.7x10⁻⁵M
<h3>CaF₂ begins precipitation when [F⁻] = 2.7x10⁻⁵M.</h3>
For BaF₂:
Ksp = 1.5x10⁻⁶ = [Ba²⁺] [F⁻]²
<em>Where [Ba²⁺] = 0.090M * {25mL / (25mL + 35mL)} = 0.0375M</em>
1.5x10⁻⁶ = [0.0375M] [F⁻]²
[F⁻]² = 4x10⁻⁵
[F⁻] = 6.3x10⁻³M
BaF₂ begins precipitation when [F⁻] = 6.3x10⁻³M
Thus, the fluoride which precipitates first is CaF₂
Answer:
c. isotope number
Explanation:
Mass Number is the sum total of mass of protons and neutrons present in the nucleus of an atom. Generally they are being used in distinguishing isotopes. E.g Carbon - 12, Carbon - 13
Atomic Number is the number of protons. Every single element has it's unique atomic number and can be used in identification purpose. E.g Carbon - 6, Hydrogen - 1.
The correct option is option C. This is the symbol that is not necessary for the identification of a nuclide.
