Silver (Ag) is the number of atoms per unit cell for each metal. Silver has a face-centred cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
The identical unit cells are defined in such a way that they take up space without touching one another. A crystal's internal 3D arrangement of atoms, molecules, or ions is known as its lattice. It consists of a large number of unit cells. Every point of the lattice is occupied by one of the three component particles.
Primitive cubic, body-centred cubic (BCC), and face-centred cubic are the three types of unit cells (FCC). The three different sorts of unit cells will be thoroughly covered in this section.
To learn more about the unit cell refer here:
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Answer:
They are both colorless, odorless, and tasteless. They have the same number of valence electrons too. And unbalanced electrons in their valence shell.
Explanation:
<u>Answer</u>:-
Name of element = Lead
Symbol ♾ of Lead = Pb
Atomic no. = 82
Atomic mass = 207.2 amu
No. of protons = 82
No. of electrons = 82
Yes, lead(Pb) is a metallic element and certainly it has 6 electron shells which means 6 energy level.
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Answer:
24.7 amu
Explanation:
An isotope is when an element can have different number of neutrons but they have same number of protons.
In order to calculate the average atomic mass with the given information do the following operations:
First change de percentages to fractional numbers, divide by 100.
I like to make a table, to organize all data and I believe is easier to understand.
65/100 = 0.65
35/100 = 0.35
% fraction
65.0 0.65
35.0 0.35
total100.0 1
Now multiply each mass with their corresponding fraction
24 (0.65) = 15.6
26 (0.35) = 9.1
% fraction uma uma
65.0 0.65 24 15.6
35.0 0.35 26 9.1
total100.0 1 24.7
Finally you add the resulting mass and the units will be in uma.
15.6+9.1 = 24.7
Therefore the average atomic mass of this element will be 24.7 uma.
Check the table in the document attached
Answer: 116 g of copper
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr =
(1hr=3600s)

of electricity deposits 63.5 g of copper.
352800 C of electricity deposits =
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles