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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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Crank

Answer:

A

Explanation:

Polypeptides such as proline and glycine are capable of making a sharp turn called a reverse turn due to the suitability and flexibility of their cyclic structure

8 0
3 years ago
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Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How many grams of Mg(OH)2 are needed to neutralize the acid in
Allushta [10]

Answer:

1.95g of Mg(OH)2 are needed

Explanation:

Mg(OH)2 reacts with HCl as follows:

Mg(OH)2 + 2 HCl → MgCl2 + 2H2O

<em>Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl</em>

To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:

<em>Moles HCl:</em>

158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl

<em>Moles Mg(OH)2:</em>

0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2

<em>Mass Mg(OH)2 -Molar mass: 58.3197g/mol-</em>

0.3350 moles Mg(OH)2 * (58.3197g / mol) =

<h3>1.95g of Mg(OH)2 are needed</h3>
3 0
3 years ago
The volume of a gas is always the same as the volume of __________.
Len [333]
The answer would be A: its container.

Gas will expand to fill whatever container it is put in.
7 0
3 years ago
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. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

5 0
3 years ago
Upvoting all answers !!
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