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Mekhanik [1.2K]
3 years ago
12

Americans combined drive about 4.0 x 109 kilometers a day and get an average of 20 miles per gallon of gasoline. For each kilogr

am of gasoline that is burned about 3 kg of carbon dioxide (CO2) are produced. With the density of gasoline as 0.93 g/cm3, how many kg of CO2 are emitted by U.S. cars into the atmosphere each minute? (1 gallon = 3.7854 L, 1 mile = 1.609 km)
A. 9.12 x 105 kg CO2/min
B. 5.47 x 107 kg CO2/min
C. 4.7 x 109 kg CO2/min
D. 4.38 x 106 kg CO2/min
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
8 0

Answer:

303,882.84649 kg\times 3=9.12\times 10^5 kg of carbon dioxide gas.

Explanation:

Average distance covered by Americans in a day= 4.0\times 10^9 km

1 day = 24 × 60 min = 1,440 min

Average distance covered by Americans in a minute= \frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km

Average mileage of the car = 20 miles/gal = 32.18 km/gal

1 mile = 1.609 km

20 miles = 20 × 1.609 km = 32.18 km

Volume of gasoline used in minute = \frac{2,777,777.78 km}{32.18 km/gal}

V=86,320.00 gal

V=86,320.00\times 3.7854 L

(1 L = 1000 mL)

V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL

Mass of 86,320.00 gallons of gasoline = m

Density of the gasoline = d = 0.93 g/cm^3=0.93 g/mL

1 mL= 1 cm^3

m=d\times V=0.93 g/mL\times 326,755,748.91 mL

m=303,882,846.49 g=303,882.84649 kg

1 kilogram of gasoline gives 3 kg of carbon dioxde gas .

Then 303,882.84649 kg of gasoline will give :

303,882.84649 kg\times 3=9.12\times 10^5 kg of carbon dioxide gas.

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While in Europe, if you drive 109 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite
MrRissso [65]

Answer:

The amount is x  =  113.3 \  dollars

Explanation:

From the question we are told that

     The  distance traveled per day is  l  =  109\ km =  \frac{109}{1.609}  =  67.74 \  mi

    The  cost of one liter is  c =  1.10 \ euros/liter = 1.10  *  1.26  = 1.36 \ dollars/liter

     The car's  gas mileage is  b =  22.0 \ mi/gal

Generally the amount of distance covered in one week is evaluated as

      z =  67.74 * 7

       z = 474.18 \  mi

The  amount of gas used in one week by the car is mathematically represented as  

       k  = \frac{ z}{ b}

=>    k  = \frac{ 474.18}{ 22}

=>      k  =  22 \ gal

converting to liters

          k =  22 *  3.78541=83.28 \ liters

Thus the amount spent on gas in one week is  

          x =  k *  c

=>      x  =  83.28 *  1.36

=>      x  =  113.3 \  dollars

 

5 0
3 years ago
) If you started with 2.3182 g of 3-nitrophthalic acid and had a 93% yield of 3-nitrophthalhydrazide, how many grams of 3-nitrop
MArishka [77]

Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.

Explanation:

The balanced chemical reaction will be :

C_8H_5NO_6+N_2H_4\rightarrow C_8H_5N_3O_4+2H_2O

moles of 3-nitrophthalic acid  = \frac{\text {given mass}}{\text {molar mass}}=\frac{2.3182g}{211.13g/mol}=0.0110mol

As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide

0.0110 moles of 3-nitrophthalic acid gives = \frac{1}{1}\times 0.0110=0.0110 mole of 3-nitrophthalhydrazide

mass of 3-nitrophthalhydrazide = moles\times {\text {molar mass}}=0.0110mol\times 177.16g/mol=1.9488g

As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = \frac{1.9488\times 93}{100}=1.8124g

Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.

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Answer:

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