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monitta
3 years ago
6

if the lightbulb receives 100 J of electrical energy, and gives off 75 energy, how much heat (thermal energy away from the light

bulb?

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
8 0

Answer:

Amount of heat energy released by light bulb = 25 joules

Explanation:

Given:

Energy receive by light bulb = 100 Joules

Energy released by light bulb as light energy = 75 Joules

Find:

Amount of heat energy released by light bulb

Computation:

We know that, energy is neither be created nor destroys

So,

Using Law of conservation of energy

Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb

100 = 75 + Amount of heat energy released by light bulb

Amount of heat energy released by light bulb = 100 - 75

Amount of heat energy released by light bulb = 25 joules

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A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
anygoal [31]

Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

<em>Where:</em>

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

<em />

The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.

3 0
2 years ago
PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
1.31 times 10^-22 / 6.6262 times 10^-34
victus00 [196]
The first answer is -.595454 the second answer is -1.9488
4 0
3 years ago
How do you find the highest energy level?
joja [24]

Answer:

The larger the number of the energy level, the farther it is from the nucleus. Electrons that are in the highest energy level are called valence electrons. Within each energy level is a volume of space where specific electrons are likely to be located.

4 0
2 years ago
Calculate the molar mass of Co2
olga nikolaevna [1]

Answer:

44.01 g/mol

Explanation:

Add each elements atomic mass. For oxygen you will do that twice because their is two oxygens.

- Hope that helps! Please let me know if you need further explanation.

8 0
2 years ago
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