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zubka84 [21]
2 years ago
8

What are the uses of hydrogen? ​

Chemistry
2 answers:
Akimi4 [234]2 years ago
8 0

<u>Following are some important uses of hydrogen:</u>

•Hydrogen is used in the synthesis of ammonia and the manufacture of nitrogenous fertilizers.

•Hydrogenation of unsaturated vegetable oils for manufacturing vanaspati fat.

•It is used in the manufacture of many organic compounds, for example, methanol.

Agata [3.3K]2 years ago
3 0

USES OF HYDROGEN

  • commercial fixation of nitrogen from the air in the Haber ammonia process hydrogenation of fats and oils

  • methanol production, in hydrodealkylation, hydrocracking, and hydrodesulphurization rocket fuel
  • welding
  • production of hydrochloric acid
  • reduction of metallic ores
  • for filling balloons (hydrogen gas much lighter than air; however it ignites easily) liquid H2 is important in cryogenics and in the study of superconductivity since its melting point is only just above absolute zero

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Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
2 years ago
PLZ HELP
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Read 2 more answers
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