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3 years ago

For the reaction, 2Cr2+ + Cl2(g) ---> 2Cr3+ + 2Cl- Ecell (standard conditions) = 1.78V

Chemistry

1 answer:

Lapatulllka [165]3 years ago

5 0

Answer:

Eºcell = -1.78 V

Explanation:

The Eº cell is an intensive property, i.e they do not depend on the quantity of material present and the desired reaction in our problem is exactly half the reverse of the one given, the Eºcell will then be the negative of the first then Eºcell is -1.78 V and the redox reaction will be non-spontaneous as opposed to the first.

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3 years ago

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If 15.00 mL of 0.0100 M Ca(IO3)2 solution are mixed with 0.500 g KI, what is the theoretical yield (in grams) of I2?

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The theoretical yield of I2 in the reaction would be 0.23 g

<h3>Theoretical yield</h3>

This refers to the stoichiometric yield of a reaction.

From the equation of the reaction:

Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O

The mole ratio of Ca(IO3)2 and I2 is 1: 6

Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100

= 0.00015 mole

Equivalent mole of I2 = 0.00015 x 6

= 0.009 mole

mass of 0.0009 I2 = 0.0009 x 253.809

= 0.23 g

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3 years ago

List the following objects in order from lowest inertia to highest inertia

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2 years ago

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Gasoline burning in an engine: Exothermic or Endothermic?

Alinara [238K]

An exothermic reaction releases heat. An endothermic reaction absorbs heat. Burning gas releases heat so it would be exothermic. Acid and water react heating the beaker would be exothermic because it releases heat from the reaction. Hope this helps! ;)

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3 years ago

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

mylen [45]

yield = 52.23 %

Explanation:

We have the following chemical reaction:

2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)

mass of aluminium = mass of bottle with aluminium pieces - bottle mass

mass of aluminium = 10.8955 - 9.8981 = 0.9974 g

mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if 2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 / 0.03694) × 100

yield = 52.23 %

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3 years ago