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Usimov [2.4K]
3 years ago
10

Described uranium (a type of nuclear fuel)​

Chemistry
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

Uranium are being used as nuclear fuel as it easily split away the atoms.

Explanation:

Uranium are generally used as fuel for nuclear fission reaction in the nuclear power plants. They use U-235 as fuel, as it can split away the atoms easily. Other than silver, uranium is more common but its availability is very low.i.e. only 0.7%. when U-235 absorbs neutron, it undergoes nuclear fission. The heavy nucleus gets splitted over two or even more lighter nuclei, along with release of kinetic energy, gamma rays and also some free electrons.

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I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.

The steps are already in the correct order.

1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.

2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.

3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.

4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.

5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).

6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.

7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.

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The _______ elements tend to lose electrons and form positive ions, while the _______ elements tend to gain electrons and form n
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Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

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3 years ago
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