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BabaBlast [244]
3 years ago
11

Electron configuration of Phosphorus

Chemistry
1 answer:
Aleks [24]3 years ago
3 0
1s^2 2s^2 2p^6 3s^2 2p^3 or the shortcut way is [Ne] 3s^2 2p^3
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34. Explain how dimensional analysis is used to solve<br> problems.
alisha [4.7K]

Answer: By understanding conversion factors and how they are related to each other

Explanation:

Dimensional Analysis is a step by step approach to solving problems in Physics, Chemistry , and Mathematics. It involves having a clear knowledge and understanding to be able to convert a given unit to another in the same dimension using  conversion factors and knowing how they are related to each other.

For instance, In Chemistry, we want to Convert 120mL to L.(note that ml stands for millilitres and ;L stands for litres)

Or first approach will be to write out the conversion factor related to our problem which is

1000ml =1L

such that 120ml = (we cross multiply))

giving us  120ml x 1L/1000ml =0.12L

This same process is applied to convert any type of dimensional analysis problems be it physics or mathematics.

7 0
3 years ago
The missing components in the table to the right
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The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...

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4 0
3 years ago
( i need help on this one )
lara [203]

Answer:

1. False

2.True

Explanation:

7 0
3 years ago
Which of the following describes a nonmetal
anastassius [24]

Answer:

b is the answer

Explanation:

non metals are not shiny, brittle, unmalleable, and are poor conductors of thermal energy and electrical current.

4 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

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where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

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Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

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The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

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3 years ago
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