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galben [10]
2 years ago
15

In an experiment magnesium ribbon was heated in air. The product was found to be heavier than the original ribbon. Potassium man

ganate 7 was on the other hand, heated in air and product formed was to be lighter. Explain the difference on the observation made
Chemistry
1 answer:
KatRina [158]2 years ago
7 0

Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

<h3>Addition and decomposition reactions</h3>

Magnesium burns in air to produce magnesium oxide as follows:

2Mg + O_2 --- > 2MgO

Potassium manganate 7 burns to produce multiple products as follows:

2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2

Thus, the MgO will be heavier than Mg. On the other hand, MnO_2 will be less heavy than KMnO_4.

More on reactions can be found here: brainly.com/question/17434463

#SPJ1

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A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a
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Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

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Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

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Heat = 150 * 4.18 * ( 31.4 - 22.3)

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Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

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