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riadik2000 [5.3K]
3 years ago
13

In Universe L , recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, qua

ntum mechanics works just as it does in our universe, except that there are four d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the third and fourth elements in the first transition series in Universe L .
Chemistry
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

See explanation

Explanation:

The third element in the first transition series is Vanadium

The fourth element in the first transition series is chromium

Given that we have four d orbitals in universe L instead of five as we have on earth;

The electronic configuration of Vanadium in universe L is;

Ar 3d3 4s2

The electronic configuration of chromium in universe L is;

[Ar] 3d4 4s2

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What is the difference between a mixture and a compound? (need an easy two sentence answer (able to explain it to a 8 your old).
lina2011 [118]
<span> A </span>mixture<span> is made from </span>two<span> or more substances that are chemically unlike</span><span> and are not chemically joined. A </span>compound<span> is a substance formed when </span>two<span> or more elements chemically react with each other to ... substances because no new substance is formed, therefore they do not </span>have<span> any fixed properties.</span>
6 0
3 years ago
Which of these actions would increase heat transfer between two objects?
klasskru [66]

Heat transfer is the phenomenon that occurs when the two objects are in the vicinity of each other and by increasing the area of their contact. Thus, option B is correct.

<h3>What is heat transfer?</h3>

Heat transfer is a process that flows the heat from one system to another, and is because of the difference in the temperature of the two objects that are part of the system.

The methods like conduction, convection, and radiation transfer the heat from the surface area to the other object. The heat gets transferred from the area of high to the low temperature.

Therefore, option B. by increasing the surface area the heat transfer increases.

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brainly.com/question/17823456

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5 0
2 years ago
There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e
irakobra [83]

80% copper (Cu)

Cu: 80 :  63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 :  63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0
2 years ago
Is na20 a stable compound, or an unstable compound?
maria [59]

Answer:

unstable

Explanation:

3 0
3 years ago
Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
Tanzania [10]

Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0
3 years ago
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