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LekaFEV [45]
3 years ago
8

In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed

even though it was shaded from the sun by a dark cloth. At first, how did Becquerel explain this result?
Chemistry
1 answer:
marshall27 [118]3 years ago
3 0
<span>In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed even though it was shaded from the sun by a dark cloth. The explanation that he gave was that the crystals were emitting light without any external light source to stimulate or excite them. He called this phenomenon as "radioactivity'". Further, more and more experiments were done to analyse actually what happened between the crystals and the light</span>
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marshall27 [118]

Answer:

Answer is in the attachment

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2 years ago
Describe two ways the material in this lesson
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Answer:

- Dalton used creativity to modify Proust's experiment and interpret the results.

- Thomson used creativity to interpret the results of the cathode ray tube experiment.

Explanation:

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3 years ago
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6.0 L of oxygen gas is at a temperature of 5K. If the temperature of the gas is lowered to 1K at constant pressure, what is the
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Answer:

THE NEW VOLUME AT 1 K IS 1.2 L

Explanation:

Using Charles' law which states that the volume of a given gas is directly proportional to its temperature provided the pressure remains constant.

Mathematically written as;

V1/T1 = V2/T2   at constant pressure

V1 = Initial volume = 6L

T1 = initial temperature = 5K

T2 = Final temperature = 1K

V2 = final volume = unknown

Re-arranging the equation by making V2 the subject of the equation, we obtain;

V2 = V1 T2 / T1

V2 = 6 * 1 / 5

V2 = 1.2 L

The new volume of the gas sample at 1 K is 1,2 L

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3 years ago
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Static electricity is caused by ..
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2 years ago
Escriba en termino de moles, de moléculas y de masa las siguientes ecuaciones a. Fe +2HCl ________ FeCl2 + H2 b. CH4 + 2O2 _____
MA_775_DIABLO [31]

Answer:

a.

  • 1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)
  • 6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)
  • 55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b.

  • 6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.
  • 1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.
  • 16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c.

  • 3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.
  • 1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.
  • 323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

Explanation:

a. Fe (s)  +2 HCl (aq) → FeCl₂ (aq)

1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)

Calculamos cuanto son dos moles de moleculas sabiendo que:

6.02×10²³ moleculas / 1 mol  . 2 mol = 1.20×10²⁴ moleculas. Entonces

6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)

Calculamos las masas molares de cada reactivo y producto

Fe = 55.85 g

HCl = 36.45 g

FeCl₂ = 126.75 g

55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b. CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.

6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.

Calculamos las masas molares:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O g = 18 g

16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c. 3 Ag (s) + 4HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2H₂O (aq)

Calculamos cuantos moleculas contienen 3 y 4 moles:

6.02×10²³  . 3 = 1.80×10²⁴ moleculas

6.02×10²³  . 4 = 2.41×10²⁴ moleculas

3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.

1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.

Calcualmos las masas molares:

Ag = 107.86 g

HNO₃ = 63 g

AgNO₃ = 169.86 g

NO = 30 g

H₂O = 18 g

323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

6 0
3 years ago
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