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Vadim26 [7]
2 years ago
9

Arrange the complexes in order of decreasing stability. Rank from most stable to least stable. To rank items as equivalent, over

lap them. [Ni(en)(H2O)4] [FeF6] [Fe(NH3)6] [Ni(en)3])
Chemistry
1 answer:
sasho [114]2 years ago
8 0

Answer: [Ni(en)3] > [Ni(en)(H2O)4] > [Fe(NH3)6] > [FeF6]

Explanation:

Generally chelating ligands stabilize the complex more than non chelated ligands and more the no of chelated ligands more the stability.

Here en (ethylenediamine) is a chelated ligand and stabilze the complex more by chelation.

And Strong field ligand (NH3) also stabilze the complex more than weak field ligand (F).

Hence F containing complex is least stable.

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Answer:

Explanation:

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3 years ago
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kvasek [131]

Answer:

Mass of sodium chloride decomposed = 24.54 g

Explanation:

Given data:

Mass of sodium chloride decomposed = ?

Mass of chlorine gas formed = 15 g

Solution:

Chemical equation:

2NaCl      →         2Na + Cl₂

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 15 g/ 71 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.

                    Cl₂            :              NaCl

                      1              :                2

                      0.21         :            2×0.21 = 0.42 mol

Mass of Sodium chloride decompose:

Mass = number of moles × molar mass

Mass = 0.42 mol × 58.44 g/mol

Mass = 24.54 g

4 0
2 years ago
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
galina1969 [7]

Answer:

\%\, yield\, \, SO_3=82.29

Explanation:

First write the balance eqation of chemical reaction:

2S(s) +3O_2(g) \rightarrow 2SO_3(g)

Remember writing any chemical reaction from its  elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen=5/32mol=0.16mol

mass of sulpher given=6gram

mole of sulpher=6/32mol=0.19mol

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with 1.5\times 0.19 i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole SO_3

1 mole of sulpher will give 2/3 mole SO_3

0.16 of sulpher will give 0.12 mole SO_3

mass of SO_3=9.6gram this is theoreical production of SO_3

and

actual production of SO_3 =7.9gram

\%\, yield   \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100

\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29

\%\, yield\, \, SO_3=82.29

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2 years ago
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Hera much hydrogen-3 wil remain after 60 years if the original sample had a mass of 100.0 g and the half-life of hydrogen-3 is 1
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12 g would be the answer to this question
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