Answer:
There are 15 total Carbon Atoms
there are 5 groups of 3
5(C3)
5x3 = 15
Both chloro and methyl groups axial
More stable
Higher concentration
Answer:
Mass = 0.00541 g
Explanation:
We will convert the larger given values in to smaller by rounding these figures.
Given data:
Mass of zinc sulfide = 43 g
Mass of oxygen = 44.2 g
Mass of zinc oxide = ?
Solution:
Chemical equation:
2ZnS + 3O₂ → 2ZnO + 3SO₂
Number of moles of ZnS:
<em>Number of moles = mass/ molar mass </em>
Number of moles = 43 g/ 97.5 g/mol
Number of moles = 0.44 mol
Number of moles of Oxygen:
<em>Number of moles = mass/ molar mass </em>
Number of moles = 44.2 g/ 32 g/mol
Number of moles = 1.4 mol
Now we will compare the moles of oxygen and zinc sulfide with zinc oxide.
ZnS : ZnO
2 : 2
0.44 : 0.44
O₂ : ZnO
3 : 2
1.4 : 2/3×1.4 =0.93
The number of moles of zinc oxide produced by ZnS are less so it will limiting reactant.
Mass of zinc oxide:
Mass = number of moles / molar mass
Mass = 0.44 mol / 81.38 g/mol
Mass = 0.00541 g
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
Answer is: 18 moles of lead(II)-nitrate.
Balanced chemical reaction:
3Pb(NO₃)₂ + 2AlCl₃ → 2Al(NO₃)₃ + 3PbCl₂.
n(Al(NO₃)₃) = 12 mol.
From chemical reaction: nAl(NO₃)₃) : n(Pb(NO₃)₂) = 2 : 3.
n(Pb(NO₃)₂) = 3 · 12 mol ÷ 2.
n(Pb(NO₃)₂) = 18 mol; amount of substance.
Al(NO₃)₃ is aluminium nitrate.
AlCl₃ is aluminium chloride.
