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.
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.
Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.
Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.
Now, we check the statements. The true ones are:
The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.
The Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.
</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
</span>
In a endothermic reaction energy from an outside source is continuously being added.
For the problem presented, the correct Lewis structure for SrO is <span>[Sr]^+2 + [O]^<span>-2. </span>I am hoping that this answer has
satisfied your query about this specific question. Also, if you have further
questions, please don’t hesitate to ask away.</span>
