Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
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
A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Answer:
The answer is 2i on right hand side.
Explanation:
We should star by checking the equation from right.
First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.
Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.
So we put 2 behind I.
Now there r 2 I on both sides.
Its simple actually.
