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zepelin [54]
3 years ago
7

Compare the amount of carbon dioxide released in one year from burning coal to power 10, 65-watt incandescent bulbs with the amo

unt released from powering 10, 13-watt compact fluorescent light (CFL) bulbs. Assume the bulbs are on four hours per day for 365 days. You will need to determine the kilowatt hours (kWh) used. First, multiply the wattage of the bulbs by the number of light bulbs to determine the total watts used in one hour. Then multiply the result by time in hours to obtain the watt hours. Next, divide the result by 1000 to obtain kilowatt hours. On average, 2.1 pounds of carbon dioxide are released for every kWh of electricity produced.
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

Incandescent bulbs: carbon dioxide released ≈ 1993 pounds

CFL bulbs: carbon dioxide released ≈ 399 pounds

Therefore, CFL bulbs release less amount of carbon dioxide as compared to Incandescent bulbs.

Explanation:

Incandescent bulbs:

Total watts = 10*65

Total watts = 650 watts

watt hours in 1 day = 650*4

watt hours in 1 day = 2600 Wh

In one year,

watt hours in 365 days = 2600*365

watt hours in 365 days = 949000 Wh

Convert into kilowatt hours

kilowatt hours in 365 days = 949000/1000

kilowatt hours in 365 days = 949 kWh

2.1 pounds of carbon dioxide is released for 1 kWh

carbon dioxide released = 949*2.1

carbon dioxide released ≈ 1993 pounds

Compact fluorescent light (CFL) bulbs:

Total watts = 10*13

Total watts = 130 watts

watt hours in 1 day = 130*4

watt hours in 1 day = 520 Wh

In one year,

watt hours in 365 days = 520*365

watt hours in 365 days = 189800 Wh

Convert into kilowatt hours

kilowatt hours in 365 days = 189800/1000

kilowatt hours in 365 days = 189.8 kWh

2.1 pounds of carbon dioxide is released for 1 kWh

carbon dioxide released = 189.8*2.1

carbon dioxide released ≈ 399 pounds

Conclusion:

Therefore, CFL bulbs release less amount of carbon dioxide (399 lb) as compared to Incandescent bulbs (1993 lb)

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Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

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First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

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\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

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The balanced chemical equation will be:

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From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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