Answer:
R aluminium = 2.63x10^-6 Ω*cm
Rcopper = 1.7 x10^-6 Ω*cm
I would use Cu as interconnections in advanced CMOS nodes.
Explanation:
the conductivity formula equals:
σ = n*g*u
n = carrier concentration
u = mobility
g = charge of carrier
The resistivity is equal to:
R = 1/σ
For the aluminium, we have:
g = 1.602x10^-19 C
R = 1/(1.98x10^23 * 1.602x10^-19 * 12 = 2.63x10^-6 Ω*cm
For copper:
R = 1/(8.5x10^22 * 1.602x10^-19 * 43.2) = 1.7 x10^-6 Ω*cm
According to the calculations found for both resistivities, I would use Cu as interconnections in advanced CMOS nodes, since copper has a lower resistivity and therefore, copper conducts electricity better.
Yes if we use the resource up before it can renew itself then it could be used up and be depleted or near depletion. The resource has to have time to come back and if we use it up to fast then the resource can not take the time it needs to come back.
Answer:
A; 
Explanation:
The best way to start solving this problem is to start with the molecule with the most atoms. Since there are 12 carbons on the left, you need 12 on the right so 12 would need to be placed in front of carbon dioxide. Also you need 22 hydrogens and in each molecule of water, there are two hydrogen molecules so you need 11 molecules of water. After balancing you find that you need 24 oxygen on the left so you place the coeffecient 12 in front of the oxygen molecule.
