What is the special state of matter that exist last at extremely high temperatures?

1 answer:

8 0

Hey there!
Usually what's taught in school is this.
Solid- Particles packed, little energy.
Liquid- Particles less packed, more energy.
Gas- High energy in particles, less particle restriction, very spread out.
But here's the extra thing,
Plasma- Examples are things like fire and lightning. Very, very, high energy.
Hopefully this helps!

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F a shell can hold a maximum of 32 electrons, what is the value of n? 2 3 4 5

Arte-miy333 [17]

Answer: 4

Explanation:

Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.

Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.

Also electrons give are 32.

The formula for number of electrons is $2n^2$ .

$2n^2=32$

$n=4$

Thus principal quantum no will be n= 4.

8 0

2 years ago

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GaryK [48]

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3 years ago

4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.

ikadub [295]

<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
[DA] Multiply/Divide [Cancel out units: $\displaystyle 0.126923 \ mol \ Au$