A 20 L sample of the gas contains 8.3 mol N₂.
According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant
<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁
<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁
___________
<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L
<em>n</em>₂ = ?; <em>V</em>₂ = 20 L
∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol
Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se
the reason is P dopant will introduce holes in the Se as P has lesser valence electron
Answer:
B. Element
Explanation:
Tungsten is an chemical element
1 significant figure, because there is no decimal after the zero the zero doesn't count.