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aivan3 [116]
3 years ago
5

Which represents the correct electron distribution of a transition element in the ground state?

Chemistry
1 answer:
Phoenix [80]3 years ago
5 0
I believe both C) and D) are correct, because both have enough electrons to show that the element has enough protons to be a transition metal, and both elements seem to have an electron configuration that exists in the ground state. We may need more information.
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Increasing temperature makes a
LenKa [72]

Answer:

High temperature increases the number of high energy collisions

Explanation:

Increasing the temperature a reaction takes place at increases the rate of reaction. At higher temperatures, particles can collide more often and with more energy, which makes the reaction take place more quickly.

5 0
3 years ago
A concentration cell consists of two Zn/Zn2+ half-cells. The concentration of Zn2+ in one of the half-cells is 2.0 M and the con
inna [77]

<u>Answer:</u> The EMF of the cell is coming when the cell having diluted concentration is getting oxidized.

<u>Explanation:</u>

We are given a cell which contains two Zn/Zn^{2+} half cells. This means that the standard electrode potential of the cell will be 0.

For a reaction to be spontaneous, the EMF of the cell must be positive. If the EMF of the cell is negative, the reaction will be non-spontaneous and will not take place.

For a reaction to be spontaneous, the diluted cell must get oxidized.

The half reaction for the given cell follows:

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}(1.0\times 10^{-3}M)+2e^-

<u>Reduction half reaction:</u>  Zn^{2+}(2.0M)+2e^-\rightarrow Zn

Net reaction:  Zn^{2+}(2.0M)\rightarrow Zn^{2+}(1.0\times 10^{-3}M)

To calculate the EMF of the cell, we use Nernst equation:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}_\text{{(diluted)}}]}{[Zn^{2+}_{\text{(concentrated)}}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Zn^{2+}_{\text{(diluted)}}] = 1.0\times 10^{-3}M

[Zn^{2+}_{\text{(concentrated)}}] = 2.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.0\times 10^{-3}M}{2.0M}

E_{cell}=0.097V

Hence, the EMF of the cell is coming when the cell having diluted concentration is getting oxidized.

6 0
3 years ago
If 2.35 L hydrogen gas react with nitrogen gas to form nitrogen trihydride, how many grams of nitrogen will be needed for the re
love history [14]

Answer:

1.47 g

Explanation:

6 0
3 years ago
Dsm-5 includes gambling disorder as an addictive disorder, along with substance use disorders. this change is considered importa
Airida [17]

Dsm-5 includes gambling disorder as an addictive disorder, along with substance use disorders. This change is considered important because it suggest that people may become addicted to behavior net just substance.

<h3>What is Gambling ? </h3>

Gambling disorder which involve repeated, problem gambling behavior. The behavior leads to problem for the individual families, and society.

The emotional and physical sign of gambling disorder are:

  • Anxiousness and Depression
  • Hopelessness
  • Lack of sleep
  • Pallor to skin
  • Gain loss of weight

<h3>What are the effects of gambling disorder ?</h3>

Gambling disorder make another kind of addiction that an individual uses to be able to deal with the situation. Start using the drugs alcohol to reduce the feeling of anxiety.

Thus from the above conclusion we can say that Dsm-5 includes gambling disorder as an addictive disorder, along with substance use disorders. This change is considered important because it suggest that people may become addicted to behavior net just substance.

Learn more about the Gambling Disorder here: brainly.com/question/28026101

#SPJ4

6 0
1 year ago
___au2s3 ___h2 → ___au ___h2s
liberstina [14]
The best way to balance an equation is to balance one atom at a time.
You start with two Au atoms on the left, so you know the coefficient of Au on the right has to be 2. So at first we get,
Au2S3 + H2 --> 2Au + H2S
Then, notice you have 3 sulfur atoms on the left, so you need three on the right.
Our equation becomes
Au2S3 + H2 --> 2Au + 3H2S
Lastly, we now have six hydrogen atoms on the right, and only two on the left, so we assign a three to the H2 on the left
Au2S3 + 3H2 --> 2Au + 3H2S Is the balanced final equation.

7 0
3 years ago
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