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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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3 years ago
A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a
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Answer: 0.20 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 1.40 M

V_1 = volume of stock solution = 72.0 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = 248 ml

1.40\times 72.0=m\times 248

m=0.41M

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.41 M

V_1 = volume of stock solution = 124 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = (124 +133) ml = 257 ml

0.41\times 124=m\times 257

m=0.20M

Thus the final concentration of the solution is 0.20 M.

8 0
3 years ago
When a barium atom loses two electrons what charge does it have
Darya [45]

Answer:

It becomes a positive ion and its radius decreases

Explanation:

As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.  

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What do we mean by basic MgO?
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5 0
3 years ago
At a temperature of 273 K, a 400.-milliliter gas sample has a pressure of 760. millimeters of mercury. If the pressure is change
KatRina [158]

Answer:

188.03K

Explanation:

The following data were obtained from the question:

T1 (initial temperature) =273K

V1 (initial volume) = 400mL

P1 (initial pressure) = 760mmHg

P2 (final pressure) = 380mmHg

V2 (final volume) = 551mL

T2 (final temperature) =?

Applying the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas can be obtained as follow:

P1V1/T1 = P2V2/T2

760 x 400/273 = 380 x 551/T2

Cross multiply to express in linear form as shown below:

760 x 400 x T2 = 273 x 380 x 551

Divide both side by 760 x 400

T2 = (273 x 380 x 551)/(760 x 400)

T2 = 188.03K.

Therefore, at a temperature of 188.03K, the gas sample will have a volume of 551mL

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