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3 years ago

Which of the following statements is consistent with the electron configuration 1s22s22p63s23p4?

Chemistry

1 answer:

spayn [35]3 years ago

6 0

Answer:

Explanation:

When finding the element based on its electronic configuration the subscripts gives the number of electrons in the element.

In the given example total number of subscripts are 16. This means the element has 16 electrons which is equal to the atomic number of the element. hence, the element is Sulphur.

Here, s, p, d, and f are orbitals. So, as per the EC given, 3p orbital has 4 electrons in it. Hence, option (a) is correct.

Since, s, p, d, and f are orbitals and not shells, subshells. Therefore, other options are incorrect.

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anygoal [31]

Hi!

The answer is breathing.

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3 years ago

How many moles of hydrogen<br> are in 3.7 moles of C8H11 NO2?

dangina [55]

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2 years ago

What does a hygrometer measure?

gogolik [260]

Answer:

humidity

Explanation:

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3 years ago

Read 2 more answers

Nuclear decay<br>27Al + He 》 39P<br>

dybincka [34]

Answer:

$_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

Explanation:

The unbalanced nuclear equation is

$^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}$

We can insert the subscripts, because these are the atomic numbers of the elements

$_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

$_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

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3 years ago

In a combustion chamber, ethane (c2h6) is burned at a rate of 9 kg/h with air that enters the combustion chamber at a rate of 17

nekit [7.7K]

The percentage of excess air used during combustion process of ethane will be 37 %.

Burning, also known as combustion, would be a high-temperature highly exothermic chemical process that occurs when an oxidant, typically atmospheric oxygen, interacts with a fuel to generate oxidized, frequently gaseous products in a mixture known as smoke.

Calculation of percentage of air .

$C_{2} H_{6} + (1-x)+a(O_{2} +3.76N_{2} )=bCO_{2} + cH_{2} O + axO_{2} + 3.76dN_{2} .$