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Allisa [31]
3 years ago
7

How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Chemistry
1 answer:
zhenek [66]3 years ago
5 0

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

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Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide
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Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

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5 0
3 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

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ΔS° = -0.050 kJ/mol.K

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Answer:

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Tcecarenko [31]

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The number of protons and neutrons.

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