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Afina-wow [57]
2 years ago
13

A website promoting the use of alternative energy vehicles and hybrid technologies claims that "A typical automobile in the USA

uses about 500 gal of gasoline every year, producing about 5 tons of carbon dioxide." To determine the truth of this statement, calculate how many tons of carbon dioxide are produced when 500.0 gallon of gasoline are combusted. Assume that the primary ingredient in gasoline is octane, C₈H₁₈(l), which has a density of 0.703?
Chemistry
1 answer:
svetlana [45]2 years ago
4 0

Answer:

The statement is correct since 500 gallon of gasolin produces 4.56 ton of carbon dioxide.

Explanation:

The combustion of gasoline follows the reation bellow:

C₈H₁₈ (l) + 17,5O₂ (g) → 8CO₂ (g) + 9H₂O (g)

This means that each mol of octane produces 8 mol of CO₂.

1 gallon ______ 4546.09 mL C₈H₁₈

500 gallon ____    x

x = 2,273,045.00 mL C₈H₁₈

1 mL gasoline ______________ 0.703 g

2,273,045.00 mL C₈H₁₈ ____  y

y = 1,597,950.635 g C₈H₁₈

1 mol C₈H₁₈ _____ 126.239 g

            y          _____  1,597,950.635 g

y = 12,966.274 mol C₈H₁₈

1 mol C₈H₁₈ ___________ 8 mol CO₂

12,966.274 mol C₈H₁₈ ____  w

w = 103,730.192 mol CO₂

1 mol CO₂ ______________ 44 g

103,730.192 mol CO₂______   z

z = 4564128.448 g CO₂

500 gallon of gasolin produces 4.56 ton of carbon dioxide.

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A chemist pours 1 mol of zinc granules into one beaker and 1 mol of zinc chloride powder into another beaker. What do the two sa
Oksana_A [137]
Answer is: <span> two samples have in common same amount of substance and same number of particles.
1) There are same amount of substance in both beakers:
n(Zn) = 1 mol.
n(ZnCl</span>₂) = 1 mol.
2) There are same number of particles (atoms, molecules, ions) in both beakers:
N(Zn) = n(Zn) · Na.
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N(ZnCl₂) = n(ZnCl₂) · Na.
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5 0
2 years ago
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The diagram below shows the different phase transitions that occur in matter. Three bars are shown labeled Solid, Liquid, and Ga
Sergio [31]

Answer:

Molecules are speeding up as boiling occurs.

Explanation:

We have three states of matter;solid liquid and gas. Substances are found in these different states of matter according to the degree of energy and velocity of its particles. Highly energetic particles moving at high velocities are found in the gaseous state. Less energetic particles moving at lesser velocities due to intermolecular forces are found in the liquid state while particles with the least degree of freedom are found in the solid state. Solid particles do not translate but can vibrate or rotate about a fixed position.

When a liquid boils, particles at the surface of the liquid acquire sufficient energy and escape the surface of the liquid. This is because, as energy is supplied in the form of heat during boiling, molecules acquire sufficient energy to speed up their molecular motion and escape the liquid surface as vapour.

6 0
2 years ago
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It’s the BOA not the dog or kangaroo
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1 year ago
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
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Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
2 years ago
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lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

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Explanation:

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a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

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          7.7 x 10⁻⁹ mol          --------------    x

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b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
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