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Ilia_Sergeevich [38]
3 years ago
5

Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochem

ical series for ligands and metals? (Select one answer. There is only one correct answer)
a. Strong o-donor ligands raise the energy of the eg orbitals of an octahedral complex.
b. Higher oxidation state metals form stronger bonds with ligands.
c. Ligands such as CO, PPhy, and bipyridine can act as T-acids.
d. The 4d and 5d transition metals are stronger Lewis acids than the 3d transition metals.
e. The electronegativity of the donor atom is the most important factor for ligands
Chemistry
1 answer:
MArishka [77]3 years ago
5 0

Answer:

Higher oxidation state metals form stronger bong with ligands

Explanation:

Ligand strength are based on oxidation number, group and its properties

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5pts. Chemistry multiple choice​
aleksandrvk [35]

Answer:

C. Xenon

Explanation:

The question only requires you to use your visual skills to identify the unknown gas. Look at the spectra of unknown gas and xenon, are they not identical? SInce they are same, the unknown sample must be xenon.

4 0
2 years ago
Read 2 more answers
The gene for tallness (T) in a pea plant is dominant over the gene for shortness (t).
astra-53 [7]

Answer:

C. 75%

Explanation:

From a heterozygous cross (Tt x Tt) the produced offspring would consist of:

  • 25% TT -it would be tall-.
  • 50% Tt -it would be tall as the gene T is dominant-.
  • 25% tt -it would not be tall-.

Thus the produced offspring that would be tall is (50% + 25%) 75%. The answer is option C.

6 0
2 years ago
Read 2 more answers
Select the correct answer.
matrenka [14]

Answer:

At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.

When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.

Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.

Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.

Explanation:

Hope This Helps Amigo!

5 0
3 years ago
If a solution has virtually virtually no hydrogen ions, what type of pH does it have? ​
GrogVix [38]

Answer:

13: if thats one of the options

Explanation:

B/c for pH to be that high you have to have alot of OH ions and almost non H+. pH won't be 0 or 2 cuz that implies ALOT of H ions. pH wont be 7 cuz that implies we have 1*10^-7 H ions.

7 0
2 years ago
Read 2 more answers
Identify the oxidizing agent and the reducing agent in the reaction. 8H +( aq) + Cr 2O 7 2–( aq) + 3SO 3 2–( aq) → 2Cr 3+( aq) +
S_A_V [24]

Answer:

SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.

Explanation:

Oxidation reaction:

3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻                  

Reduction reaction:

Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)

Now, adding the oxidation and the reduction reactions we get the full net reaction:

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.

<u>Therefore, SO₃²⁻ is the reducing agent. </u>

And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.

<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>

3 0
2 years ago
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