Answer:
The feed ratio (liters 20% solution/liter 60% solution) is 3,08
Explanation:
In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.
100 kg of 20% solution are 100kg/1,139 kg/L = <em>87,8 L </em>
100kg×20wt% = 20 kg H₂SO₄. In moles:
20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ <em>203,9 mol</em>
The final molarity 4,00M comes from:
<em>(1)</em>
Where X moles and Y liters comes from 60,0 wt% H₂SO₄
100 L of 60,0 wt% H₂SO₄ are:
100L××
×
= <em>0,9164 kmolH₂SO₄ ≡ 916,4 moles</em>
That means:
X/Y = 916,4/100 = 9,164 <em>(2)</em>
Replacing (2) in (1):
Y(liters of 60,0 wt% H₂SO₄) = <em>28,52 L</em>
Thus, feed ratio (liters 20% solution/liter 60% solution):
87,8L/28,52L = <em>3,08</em>
I hope it helps!
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility, = 0.14 m²/V-s
Hole mobility, = 0.023 m²/V-s
Now,
σ =
or
σ =
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³