Explanation:
An electron cloud is the region of space that surrounds the nucleus of an atom under the nucleus of the negatively charged sub-atomic particles.
- An atoms is made up of some particles.
- The proton is positively charged and it is located in the nucleus of an atom.
- The neutrons do not carry any charges and occupies the nucleus with protons.
- Electrons are the negatively charged particles outside the nucleus.
- The electron cloud is portion of an atom with greatest probability of finding an electron.
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Electron cloud brainly.com/question/1832385
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Q = mCΔT
Q is heat in joules, m is mass, C is specific heat, and delta T is change in temp
2099 J = (40.27g)(C)(148.5 - 24.8) = .421 J / gram K
Answer:
A ball of ice blasted into outer space
Explanation:
It is a ball of ice that is going so fast it leaves a "tail"
Answer:
To find the average atomic mass, you take a certain number of atoms, find the total mass of each isotope, and then divide the total mass of all the atoms by the total number of atoms. Assume that you have, say, 10 000 atoms of carbon
Explanation:
Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:


a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;


density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure

where;
r = 144


a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴


density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.