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bija089 [108]
3 years ago
12

Explain why it is important that there be no gaps around the edges of the filter paper during a vacuum filtration

Chemistry
1 answer:
lesantik [10]3 years ago
4 0

Vacuum filtration is more efficient and faster than filtration by the gravity method. But the set up of the vacuum filtration must be precise. Vacuum filtration is done under very reduced pressure hence there should be no gaps around the edges of the filter paper.   For a vacuum to be effective there must be no hole or gap in the equipment. Under vacuum conditions air is removed and the pressure is lowered. If there is gap then air would be continuously supplied and it would be difficult to remove air completely.



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If there are 2.54cm in one inch how many Inches are in 19.9​
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I think the answer is 50.546

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What kind of plant growth hormone to regular plant growth. We dissolve 5.0 mg of the solution into 2.0 kg of water. What is the
In-s [12.5K]

Answer:

The dosage of the solution is 2.5 parts per million.

Explanation:

The solvent proportion in parts per million is equal to the ratio of solution in miligrams to ratio of water in miligrams multiplied by a million. (A kilogram is equivalent to a million miligrams) That is:

p.p.m. = \frac{5\,mg}{2000000\,mg}\times 1000000

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3 years ago
Which solute produces the highest boiling point in a 0.15 m aqueous solution?
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ΔT = Kb * m * i

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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
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