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3 years ago

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The statement that "the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allow

ed by the Pauli principle in a particular set of degenerate orbitals" is known as: A. the quantum model B. Hund's rule C. the aufbau principle D. Heisenberg uncertainty principle E. the Pauli exclusion principle

1 answer:

lutik1710 [3]3 years ago

7 0

Answer:

B. Hund's rule

Explanation:

Hund's rule -

According to Hund's rule ,

As, the electrons are negatively charged and hence , like poles repel , so they repel each other in order to stabilizes themselves and hence , the electron firstly occupies a vacant orbital , before actually pairing up , so as to reduce repulsion .

The rule of maximum multiplicity states that ,

The term with the lowest energy is the one with the highest value of the spin multiplicity .

hence , the statement given in the question is about Hund's rule .

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In an exothermic reaction, the energy of the reactants is greater than the energy of

mojhsa [17]

Answer:

H

Explanation:

If the energy content in the reactants is higher than the products, that means that the reaction must be giving away energy. So, in other words, chemical energy from the reactants must be released as heat.

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2 years ago

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What

Fynjy0 [20]

Answer:

$l=3.5*10^{-10}m$

Explanation:

From the question we are told that:

1st Energy $E_1=12eV=4(3eV)$

2nd Energy $E_2=27eV=9(3eV)$

3rd Energy $E_3=48eV=16(3eV)$

Generally the equation for Energy E for electron in one dimensional box at ground state $E_0$ is mathematically given by

$E_0=\frac{h}{8ml^2}$

$E_0=\frac{h}{8ml^2}$

Therefore Length a is mathematically given as

$l=\sqrt{\frac{h^2}{8mE_0} }$

$l=\sqrt{\frac{(6.625*10^{-34})^2}{8(9.1*19^{-31}{(3eV(1.6*10^{-19}))}}$

$l=3.5*10^{-10}m$

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2 years ago

_____ is the process of linking smaller molecules to form long chains of higher molecular weight.

Leya [2.2K]

What is “Polymerization” is your answer

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3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

trapecia [35]

Answer:

None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

Explanation:

Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.

Before balanced Left side.

Cl-2

O-8

H-2

Before balanced right side.

H-1

Cl-1

O-3

That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.

(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)

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2 years ago