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3 years ago

Use the successive ionization energies for this unknown element to identify the family it belongs to.

Chemistry

1 answer:

NNADVOKAT [17]3 years ago

4 0

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low ionization energies for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

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Sodium thiosulfate, Na 2S 2O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction

Tatiana [17]

Answer: Reducing agent in the given reaction is $S_{2}O^{2-}_{3}$ .

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

In the given reaction, oxidation state of sulfur in $S_{2}O^{2-}_{3}$ is +2 and $I_{2}(aq)$ has 0 oxidation state.

In $S_{4}O^{2-}_{6}(aq)$ oxidation state of S is 2.5 and in $2I^{-}(aq)$ oxidation state of I is -1.

Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

Thus, we can conclude that reducing agent in the given reaction is $S_{2}O^{2-}_{3}$ .

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3 years ago

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Answer:

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Explanation:

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2 years ago

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Q 9 The table lists the steps to clean water for drinking purpose. 1. Adding chlorine tablets to the water.2. Pouring the water

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Answer:

1 and 2

Explanation:

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3 years ago

Christy is a forensic scientist. She is given

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Answer:

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Explanation:

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4 years ago

For double-helix formation, DG can be measured to be 2 54 kJ mol 1 ( 2 13 kcal mol 1 ) at pH 7.0 in 1 M NaCl at 25 8C (298 K). T

levacccp [35]

Answer:

ΔS = -661.0J/mol is the entropy change for the system

ΔS = -842J/mol.K is the entropy change for the surroundings

Explanation:

From the relationship between ΔG, T, ΔH and ΔS,

Mathematically, ΔG = ΔH - TΔS

TΔS = ΔH - ΔS

ΔS = ΔH - ΔS / T

but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)

plugging into the equation,

ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298

ΔS = -0.6610KJ/mol or in J.mol

ΔS = -661.0J/mol is the entropy change for the system

For entropy change for the surroundings = ΔS = ΔH/T

ΔS = -251 kJ/mol / 298K

ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings

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3 years ago