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Finger [1]
3 years ago
11

In the diagram below, particles of the substance are moving from the liquid phase to the gas phase at the same rate as they move

from the gas phase to the liquid phase. mc003-1.jpg The gas and liquid are at equilibrium. a high vapor pressure. a low vapor pressure. zero vapor pressure.
Chemistry
2 answers:
andrezito [222]3 years ago
6 0
The correct option is this: THE GAS AND LIQUID ARE AT EQUILIBRIUM.
A chemical system is said to be in equilibrium if the forward and the backward reactions are equal. At equilibrium, both the reactants and the products are present in concentrations which have no tendency to change with time.
harkovskaia [24]3 years ago
3 0

The gas and liquid are at equilibrium.

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4 years ago
If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form
sergiy2304 [10]

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3O_{2}⇒ 2 Fe{2}O{3}

From the given weight of iron to be used for the production of Fe{2}O{3}, number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

  Putting the values in equation  

Number of moles =  10 gm  ÷ 55.845 gm/mole

                               =  0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

 So,  2÷ 4 = x ÷ 0.179

     2/4 = x/ 0.179

    2 × 0.179 = 4x

     2 × 0.179 / 4 = x

  x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

            =  159.69 grams ×  0.0895

             = 14.292 grams of Fe2O3 formed.

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Write a hypothesis that answers the lesson
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Answer:

the reaction will come to a halt and the other reactant will still be present.

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