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Nastasia [14]
3 years ago
9

In a certain city, electricity costs $ 0.18 per kW ⋅ h . What is the annual cost for electricity to power a lamppost for 5.00 h

per day with a 100. W incandescent light bulb versus an energy efficient 25 W fluorescent bulb that produces the same amount of light? Assume 1 year = 365 days .
Chemistry
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer:

The cost of electricity  for 100 W power bulb = $ 32.85

Cost of electricity for 0.025 W  fluorescent  bulb =  $ 8.2125

Explanation:

Cost of electricity = $ 0.18 per KW-H

Time = 5 hour per day

Bulb power = 100 W = 0.1 KW

Fluorescent bulb power = 25 W = 0.025 KW

(a) Cost of electricity for 100 W power bulb

0.1 × 5 × 365 × 0.18 =  $ 32.85

(b) Cost of electricity for 0.025 W  fluorescent  bulb

0.025 × 5 × 365 × 0.18 = $ 8.2125

Therefore the cost of electricity  for 100 W power bulb = $ 32.85

Cost of electricity for 0.025 W  fluorescent  bulb =  $ 8.2125

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Answer:

um to small

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7 0
3 years ago
How do you solve this ??
ICE Princess25 [194]

Answer:

Option D. 400 mmHg

Explanation:

The following data were obtained from the question:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total pressure = 10³ mmHg

Partial pressure of He =.?

Next, we shall determine the total number of mole in the reaction vessel.

This can be obtained as follow:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total mole =?

Total mole = nHe + nNe

Total mole = 0.04 + 0.06

Total mole = 0.1

Next, we shall determine the mole fraction of He.

This can be obtained as follow:

Mole fraction = mole of gas /total mole

Mole of He (nHe) = 0.04 mole

Total mole = 0.1

Mole fraction of He =.?

Mole fraction of He = nHe/total mole

Mole fraction of He = 0.04/0.1

Mole fraction of He = 0.4

Finally, we shall determine the partial pressure of He as follow:

Partial pressure = mole fraction x total pressure

Mole fraction of He = 0.4

Total pressure = 10³ mmHg

Partial pressure of He =.?

Partial pressure of He = 0.4 x 10³

Partial pressure of He = 400 mmHg.

Therefore, the partial pressure of He is 400 mmHg.

8 0
2 years ago
Hydrochloric acid reacts faster with powdered zinc than with an equal mass of zinc strips because the greater the surface area o
inysia [295]

Answer:

increases the frequency of particle collisions

Explanation:

One factor upon which the rate of reaction depends is the surface area of reactants.

According to the collision theory, reactions occur when reactant particles having the required (activation) energy collide with each other, this collision is inelastic. However, collision of particles having energies less than the activation energy results in elastic collisions and no chemical reaction.

The more the exposed surface area of reactants, the greater the number of particles that come into contact with each other and the more the chances of frequent effective collisions that lead to reaction.

Thus, powdered zinc reacts faster with hydrochloric acid than zinc strips

7 0
3 years ago
PLEASE HELP ME !!!!!!
Leviafan [203]

Answer: b

Explanation:

By adding heat you are adding more energy

5 0
3 years ago
Read 2 more answers
Initially, a particular sample has a total mass of 200 grams and contains 128 x 1010 radioactive nuclei. These radioactive nucle
lubasha [3.4K]

Explanation:

Formula to calculate how many particles are left is as follows.

              N = N_{0} (\frac{1}{2})^{l}

where,     N_{0} = number of initial particles

                              l = number of half lives

As it is given that number of initial particles is 128 \times 10^{10} and number of half-lives is 3.

Hence, putting the given values into the above equation as follows.

               N = N_{0} (\frac{1}{2})^{l}

                    = 128 \times 10^{10}(\frac{1}{2})^{3}

                    = 16 \times 10^{10}

or,                    1.6 \times 10^{11}

Thus, we can conclude that 1.6 \times 10^{11} particles of radioactive nuclei remain in the given sample.

In five hours we've gone through 5 half lives so the answer is:

particles

8 0
3 years ago
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