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2 years ago

One d orbital can hold ___ electrons. A. 2 B. 6 C. 10 D. 14

2 answers:

8 0

Correct option is D)
Maximum number of electrons held in the d orbitals is 10.
There are 5 sub-levels in a d orbital, each one can have a maximum 2 electrons.
Hence, 5 sub-levels can have a maximum of 10 electrons.
Thus, s, p, d and f orbitals can accommodate a maximum of 2, 6, 10 an 14 electrons.

yaroslaw [1]2 years ago

8 0

${ \red{ \tt{option \: (c)}}}$

Explanation:

One d orbital can hold <u>1</u><u>0</u> electrons.

The maximum number of electrons which can be hold by one d orbital is 10 electrons. It can hold less than 10 electrons in it's orbit but not more than 10 electrons. It's impossible to hold and never possible to hold more than 10 electrons in d orbital.

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According to the Arrhenius concept, an acid is a substance that ________. Group of answer choices can accept a pair of electrons

Schach [20]

Answer: an increase in the concentration of $H^+$ in aqueous solutions and is capable of donating one or more $H^+$

Explanation:

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions $(OH^-)$ when dissolved in water and an acid is defined as a substance which donates hydrogen ions $(H^+)$ in water.

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

Thus According to the Arrhenius concept, an acid is a substance that causes an increase in the concentration of $H^+$ in aqueous solutions and is capable of donating one or more $H^+$

5 0

4 years ago

On the periodic table, in which period is copper?

Firdavs [7]

Answer:

At the top of Group 11 above silver and gold.

Period 4

Explanation:

3 0

3 years ago

Based on the information in the passage, what is true of gases?

luda_lava [24]

1: True
2: True
3: False
4: False

(Question 2 might not be true, not sure)

7 0

3 years ago

Which sample contains the largest number of oxygen atoms? Select one: a. 8.0 g of carbon dioxide b. 8.0 g of potassium chlorate

adell [148]

The sample with the largest number of oxygen atoms will be calcium perchlorate.

<h3>Number of atoms in a compound</h3>

Since we are not looking at the number of moles, the mass of the compounds has no bearing on the number of atoms of oxygen.

The chemical formula for carbon dioxide is $CO_2$ . Thus, it has 2 atoms of oxygen.
The chemical formula for potassium chlorate is $KClO_3$ . Thus, it has 3 oxygen atoms.
The chemical formula for calcium perchlorate is $Ca(ClO_4)_2$ . Thus, it has 8 atoms of oxygen.
The chemical formula for sodium hydroxide is NaOH. Thus, it has 1 atom of oxygen.

Therefore, the compound with the largest number of oxygen atoms is calcium perchlorate.

More on the number of atoms in compounds can be found here: brainly.com/question/1686912

#SPJ1

5 0

2 years ago

Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below,

Zielflug [23.3K]

The given reaction is

$C _{(Diamond)}\rightarrow C_{(Graphite)}$

An element can exist in 2 or more different forms which have totally different chemical and physical properties. They are known an allotropes.

Diamond and Graphite are allotropic forms of carbon. In the given reaction, diamond is changing to graphite and we have to find out the standard free energy of this reaction. This will help us to find out whether this reaction is spontaneous at 298 K or not.

The following data is needed for calculations which is taken from standard reference table.

$H_{f}^{0}(diamond)= 1.895 kJ/mol$

$H_{f}^{0}(graphite)= 0$

$S^{0}(diamond)=2.337 J/mol-K$

$S^{0}(graphite)=5.740 J/mol-K$

Step 1: Find ΔH⁰ rxn for the given reaction.

The formula to calculate ΔH⁰ rxn is given below.

$\bigtriangleup H^{0}_{rxn}= H_{f}(product)- H_{f}( reactant)$

We have graphite on product side and diamond on reactant side.

Therefore, $\bigtriangleup H^{0}_{rxn}= H_{f}(graphite)- H_{f}(diamond)$

Let us plug in the values given above.

$\bigtriangleup H^{0}_{rxn}= 0 - 1.895 kJ/mol$

$\bigtriangleup H^{0}_{rxn}= - 1.895 kJ/mol$

ΔH⁰ rxn for the given reaction is -1.895 kJ/mol

Step 2 : Find ΔS⁰ rxn for the given reaction.

The formula to calculate ΔS⁰ rxn is

$\bigtriangleup S^{0}_{rxn}= S^{0}(product)- S^{0}( reactant)$

$\bigtriangleup S^{0}_{rxn}= S^{0}(graphite)- S^{0}(diamond)$

$\bigtriangleup S^{0}_{rxn}= (5.740J/mol.K) - (2.337 J/mol.K)$

$\bigtriangleup S^{0}_{rxn}= 3.403 J/mol.K$

Let us convert this to kJ.

$\frac{3.403J}{mol.K}\times \frac{1 kJ}{1000J}= 3.403\times 10^{-3}kJ/mol.K$

ΔS⁰ rxn for the given reaction is 3.403 x 10⁻³ kJ/mol.K

Step 3: Find standard free energy ΔG⁰ rxn.

ΔG⁰ rxn for the given reaction is calculated as

$\bigtriangleup G^{0}_{rxn}= \bigtriangleup H^{0}_{rxn}- T\times \bigtriangleup S^{0}_{rxn}$

We have T = 298 K. Let us plug in the calculated values of ΔH⁰ rxn and ΔS⁰ rxn.

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [(298K)\times 3.403\times 10^{-3}kJ/mol.K]$

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [1.014 kJ/mol]$

$\bigtriangleup G^{0}_{rxn}= - 2.909 kJ/mol$

The standard free energy change for the given reaction is -2.909 kJ/mol

The negative value of delta G⁰ suggests that the given reaction is spontaneous at room temperature. That means diamond will slowly convert to graphite. The speed of this reaction is extremely slow, but yet the reaction is taking place. So over a period of time diamond will become graphite.

Therefore "Diamond is forever" is not true as it is going to get converted to graphite.

7 0

4 years ago