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Bad White [126]
3 years ago
8

When a magnesium ribbon is heated in air the product former is heavier.on the other hand when potassium manganate(VII) is heated

in air the product formed is lighter.Explian the difference iin observation
Chemistry
1 answer:
vesna_86 [32]3 years ago
4 0

Answer: On heating, Magnesium forms its oxide; while potassium manganate(VII) decomposes

Explanation:

Magnesium Mg, on heating forms Magnesium oxide

2Mg(s) + O2(g) --> 2MgO

Potassium permanganate KMnO4, on heating decomposes to potassium manganate K2MnO4, manganese dioxide MnO2, and Oxygen gas O2.

2KMnO4 --> K2MnO4 + MnO2 + O2

The difference in observation is that, on heating, Magnesium forms its OXIDE as product; while potassium manganate(VII) decomposes, giving OFF most of its constituents and reducing its weight.

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3 years ago
How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
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Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>

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We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

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