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Cloud [144]
2 years ago
11

I need help w this..

Chemistry
1 answer:
Mariana [72]2 years ago
3 0
Evaporation condensation transpiration precipitation
Emily, come to papa
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What happens when a catalyst is used in a chemical reaction
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3 years ago
Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4
kozerog [31]

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:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters} <em>(1)</em>

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = <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!

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