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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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USPshnik [31]

Answer:

719.83°C

Explanation:

The heat that the sample of Zinc gives is equal to the heat that water is absorbing. That is:

C(Zn) * m(Zn) * ΔT(Zn) = C(H2O) * m(H2O) * ΔT(H2O)

<em>Where:</em>

<em>C is specific heat (Zn: 0.390J/g°C; H2O: 4.184J/g°C)</em>

<em>m is mass (Zn: 2.50g; H2O: 65.0g)</em>

<em>ΔT (Zn: ?; H2O: (22.5°C - 20.0°C = 2.50°C)</em>

<em />

Replacing:

0.390J/g°C * 2.50g * ΔT(Zn) = 4.184J/g°C * 65.0g * 2.50

ΔT(Zn) = 697.33°C

As final temperature of Zn is 22.50°C, initial temperature is:

Initial temperature: 697.33°C + 22.50°C

719.83°C

<em />

3 0
3 years ago
Which of these takes place when a chemical change occurs?
zmey [24]

Answer:

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.

Explanation:

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The ___ is the short form that chemists use to identify an element
bonufazy [111]

Answer:

Chemical Symbols

Explanation:

Im not positive on this one so dont rely on me lol

8 0
3 years ago
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The percentage of water vapor present in the air compared to that required for saturation is the ____. a. mixing ratio b. relati
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Answer:

B. Relative Humidity

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3 years ago
What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?
dusya [7]
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

          6.86 = -㏒ Ka 

∴Ka = 1.38 x 10^-7

by using ICE table:

               H2PO4- →  H+  + HPO4
initial      0.4 m            0         0

change     -X                +X       +X

Equ       (0.4-X)               X          X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X)   by solving for X

∴X = 2.3x 10^-4 

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

        = -㏒ (2.3 x 10^-4)
 ∴PH  =  3.6

3 0
3 years ago
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