Explanation:
96.485 columbs=1 faraday will
deposit 64/2g= 32 g cu ion
therfore it will require
96,485 ×2/32 =? coulombs or 1/16 of
Faraday= 1 / 16 mole of electrons .
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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Carbon dioxide
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Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>
Answer: Summary of Common Properties
- High ionization energies.
- High electronegativities.
- Poor thermal conductors.
- Poor electrical conductors.
- Brittle solids—not malleable or ductile.
- Little or no metallic luster.
- Gain electrons easily.
- Dull, not metallic-shiny, although they may be colorful.
Explanation:
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