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Delicious77 [7]
4 years ago
13

A process called gaseous diffusion is often used to separate isotopes of uranium-that is

Chemistry
1 answer:
algol [13]4 years ago
5 0

Answer:

<h2>uranium-235 (²³⁵U) and uranium-238 (²³⁸U)</h2>

Explanation:

The gaseous diffusion process utilizes uranium hexafluoride, UF₆, because although it is a solid at room temperature it is easily vaporized. [1] UF6 is not only convenient for its volatility, but also due to the fact that fluorine only consists of the isotope ¹⁹F, meaning the difference in molecular weights for UF6 are purely reliant on 235U and 238U.Here arises another problem however, for the masses of the two uranium isotopes are so nearly equal there is very little separation of 235UF6 and 238UF6 with one pass through a diffuser.Therefore a cascade process is needed to obtain any measurable amount of enrichment. In a cascade the feed stream at diffuser 1 is the UF6 prior to enrichment (meaning it will contain 0.711% 235U and 99.289% 238U) and marks the start of the cascade. There will be hundreds to thousands of diffusers on the upward or enriching side as well as on the downward or depleted side. The slightly enriched UF6 is sent up the cascade process to the next diffuser where it will be enriched again. The slightly depleted UF6 will be sent downward through the cascade where it will also be enriched again. In this way, the enriched uranium keeps getting enriched and sent onward, and the depleted uranium also gets enriched and sent onward. The depleted uranium always gets sent downward where it will eventually be ejected from the downward stream as depleted uranium.

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Predict the products and then balance the equation
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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

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(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

    Question 1:

  • <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>

    Question 2:

  • <u>0.201 M</u>

Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

  • V = 0.700 liter
  • M = 0.290 mol/liter
  • M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

  • Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

  • Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0
3 years ago
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