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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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Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
Sheela wants to make a computer model of a land ecosystem. Part of her model simulates the chemical processes of photosynthesis
shusha [124]
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5 0
3 years ago
Read 2 more answers
True or false natural elements don't reflect light.
Burka [1]
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3 0
3 years ago
The equilibrium constant, Kp, for the following reaction is 0.497 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of t
Anestetic [448]

<u>Answer:</u> The equilibrium partial pressure of chlorine gas is 0.360 atm

<u>Explanation:</u>

For the given chemical equation:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)

The expression of K_p for above reaction follows:

K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}

We are given:

K_p=0.497\\p_{PCl_3}=0.651atm\\p_{PCl_5}=0.471atm

Putting values in above equation, we get:

0.497=\frac{p_{Cl_2}\times 0.651}{0.471}\\\\p_{Cl_2}=0.360atm

Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm

5 0
3 years ago
How many grams of ca are needed to react completely with 2.20 L of a 4.50 m hcl solution
Dmitry_Shevchenko [17]
Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

5 0
3 years ago
Read 2 more answers
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