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Aloiza [94]
3 years ago
12

Given the unbalanced equation:

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

3 is the coefficient in front of the CaSO_{4}

The smallest whole-number coefficient is 2.

Explanation:

The balanced chemical equation is as follows.

Al_{2}(SO_{4})_{3}(aq)+3Ca(OH)_{2}(s) \rightarrow 2Al(OH)_{3}(s)+3CaSO_{4}(s)

3 is the coefficient in front of the CaSO_{4}

The smallest whole-number coefficient is 2.

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PIT_PIT [208]

Answer:

0.000696

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1 year ago
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rodikova [14]

Answer: C

Explanation:

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3 years ago
Determine if the electron configurations are valid or invalid
Hunter-Best [27]

Answer:

1) A (valid)

2) B (Invalid)

3) B (Invalid)

4) A (valid)

5) B (Invalid)

6) B (Invalid)

Explanation:

Electronic configuration:

The electronic configuration is the arrangnment of electrons in orbitals (s, p, d, f, g, h so on) according to quantum  number and Aufbau rule.

An atom of element have orbitals, and electrons are revolving in it.

There are quantum numbers that describe the number of electron in specific shell.

The Principle quantum number is denoted by "n" and can not be "0".

These "n" is K, L, M, N and electrons are distribution in it.

K =2

L= 8

M= 18

According to Aufbau electron occupied in the lowest energy level before occupying the highest energy level.

According to this the electronic configuration is

1s², 2s², 2p⁶ and so on....

So,

"s" orbital accommodate maximum of 2 electrons

"p" orbital accommodate maximum of 6 electrons

"d" orbital accommodate maximum of 10 electrons

"f" orbital accommodate maximum of 14 electrons

and the coefficient 1, 2, 3, 4 are energy level of orbitals.

The 1st orbital is 1s and it only can occupy 2 electrons the next coming electron goes into 2s that also can occupy 2 electrons and the further electrons will go to 2p orbital.

So keeping in mind all above details we investigate all option one by one

1.  A (valid)

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵

This is valid option. number of electrons are rightly accommodated in orbitals. Also 3d is higher energy level than 4p.

______________

2.  B (Invalid)

1s², 2s², 2p⁶, 3s², 3p⁷

This is invalid option. number of electrons are not rightly accommodated in orbitals as "p" orbital can only accommodate 6 electrons and not more than 6.

________________

3.  B (Invalid)

[Xe], 7s², 6d¹, 5f⁸

This is Invalid option. Number of electrons are not rightly accommodated in orbitals. There should 5f orbital before 6d.

______________

4.  A (valid)

[Kr] 5s², 4d¹⁰, 5p⁵

This is valid option. number of electrons are rightly accommodated in orbitals.

______________

5.  B (Invalid)

[He]--

This is invalid option. There is no exact explanation of electron distribution in orbitals.

______________

6.  B (Invalid)

1s², 2s², 2p⁶, 3s³, 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s--

This is invalid option. number of electrons are not rightly accommodated in orbitals. As "s" can accommodate only 2 electrons not 3 electrons. also there is no details of 5s

______________

4 0
3 years ago
3/4 / 100 as a decimal and a percentage?​
Rufina [12.5K]
75 present
75.00 decimal
4 0
3 years ago
Xenon tetrafluoride has two sets of lone pairs of electrons. What should be the relative positions of the two sets of lone pairs
Zielflug [23.3K]

Answer: Two equatorial positions so as to minimize the repulsion.

Explanation:

Formula used for calculating number of electrons :

\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom  = 8

N = number of monovalent atoms bonded to central atom = 4

C = charge of cation  = 0

A = charge of anion  = 0

Now we have to determine the hybridization of the XeF_4   molecule.

\frac{1}{2}[8+4-0+0]=6

Bond pair electrons = 4

Lone pair electrons = 2

The number of electrons are 6 that means the hybridization will be sp^3d^2 and the electronic geometry of the molecule will be octahedral.

The molecular geometry will be square planar as two alternate positio ns will be occupied by lone pair of electrons so as to minimize the repulsion.

4 0
3 years ago
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