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barxatty [35]
3 years ago
7

For a given reaction, ΔH=78.7kJ/mol, and the reaction is spontaneous at temperatures above the crossover temperature, 461K. The

value of ΔS= _______ J/mol⋅K, assuming that ΔH and ΔS do not vary with temperature.
Chemistry
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

171 J/mol.K

Explanation:

The spontaneity of a reaction depends on the Gibbs free energy (ΔG). If ΔG < 0, the reaction is spontaneous. The Gibbs free energy is related to the enthalpy (ΔH) and the entropy (ΔS) of the reaction through the following expression:

ΔG = ΔH - T.ΔS

where,

T is the absolute temperature

If ΔG < 0,

ΔH - T.ΔS < 0

ΔH < T.ΔS

ΔS > ΔH/T = (78.7 × 10³ J/mol)/461 K = 171 J/mol.K

ΔS must be at least 171 J/mol.K for the reaction to be spontaneous above 461 K.

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Answer:

Less

Explanation:

Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:

  • The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
  • The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand

A partial spectrochemical series listing of ligands from small Δ to large Δ:

I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO

Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.

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4 years ago
The energy of colored light increases in the order red, yellow, green, blue, violet.List the metallic elements used in the flame
Vinil7 [7]

Answer:

Explanation:

Flame test:

The metals ions can be detected through the flame test. Different ions gives different colors when heated on flame. Tom perform the flame test following steps should follow:

1. Dip a wire loop in the solution of compound which is going to be tested.

2. After dipping put the loop of wire on bunsen burner flame.

3. Observe the color of flame.

4. Record the flame color produce by compound

Color produce by metals:

Red = Lithium, zirconium, strontium, mercury, Rubidium (red violet)

Orange-red = calcium

Yellow = sodium, iron (brownish yellow)

Green = green

Blue = cesium. arsenic, copper, tantalum, indium, lead

Violet = potassium (lilac)

3 0
4 years ago
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1. MgCl2 + Zn -> Mg + ZnCl2

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3. 2Cd + O2 -> 2CdO

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5. Zn + H2SO4 -> ZnSO4 + H2

6.2KBr + Cl2 -> 2KCl + Br2

7. AgNO3 + Na -> NaNO3 + Ag

8. 2NaCl + F2 -> 2NaF + Cl2

9. AgNO3 + Mg(NO3)2 + Ag

10. Ni + H2SO4 -> NiSO4 + H2

11. Al + K2SO4

12. FeCl3 + 3Mn-> 3MnCl + Fe

13. 2Na + 2H2O -> 2NaOH + H2

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3 0
3 years ago
Theoretically, how far away from the Earth do you have to go to feel zero gravitational force from the Earth?
EastWind [94]

Answer:

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Hope it helps

Explanation:

5 0
3 years ago
Read 2 more answers
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %
11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

<em>Note: The question is stated more clearly below:</em>

<em>Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.</em>

<em>What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?</em>

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

6 0
3 years ago
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