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fredd [130]
2 years ago
11

Automobiles are often implicated as contributors to global warming because they are a source of the greenhouse gas CO2. How many

pounds of CO2 would your car release in a year if it was driven 170. miles per week? Gasoline is a complex mixture of hydrocarbons. In your calculations, assume that gasoline is isooctane (molecular formula C8H18) and that it is burned completely to CO2 and H2O in the engine of your car. Also assume that the car averages 27.5 miles per gallon and that the density of isooctane is 0.692 g cm-3.
Chemistry
1 answer:
kondor19780726 [428]2 years ago
7 0

Answer:

5,747.82 pounds of CO_2 would your car release in a year if it was driven 170 miles per week.

Explanation:

Mileage of the car = 27.5 miles/gal

Distance driven by car in week = 170 miles

Distance driven by car in a day= \frac{170 miles}{7}=24.286 mile

Volume of gasoline used in a day : V

\frac{24.286 mile}{27.5 miles/gal}=0.8831 gal=0.8831\times 3785.41 cm^3=3,342.96 cm^3

1 gal = 3785.41 mL = 3785.41 cm^3

Density of the gasoline = d= 0.692 g/cm^3

Mass of the gasoline used in a day = m

m=d\times V=0.692 g/cm^3\times 3,342.96 cm^3=2,313.33 g

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

Moles of gasoline ( isooctane):

=\frac{2,313.33 g}{114 g/mol}=20.292 mol

According to reaction , 2 moles of isooctane gives 16 moles of carbon dioxide gas.Then 20.292 mole isooctane will give :

\frac{16}{2}\times 20.292 mol=162.340 mol of carbon dioxide gas

Mass of 162.340 moles of carbon dioxide gas :

162.340 mol × 44 g/mol = 7,142.91 g

Mass of carbon dioxide gas produced in a day =  7,142.91 g

1 year = 365 days

Mass of carbon dioxide gas produced in a 365 days:

= 365 × 7,142.91 g = 2,607,163.494 g

1 pound = 453.592 g

2,607,163.494 g=\frac{ 2,607,163.494}{453.592} pounds=5,747.82 pounds

5,747.82 pounds of CO_2 would your car release in a year if it was driven 170 miles per week.

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snow_tiger [21]

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<h3>Further explanation</h3>

Hydrate is a compound that binds water (H₂O), usually in the form of crystals/ solids

If these compounds are dissolved in water or heated, the hydrates can decompose:

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The formula for the hydrated compound contains: YH2O

The mole ratio shows the ratio of the coefficients of the hydrate compound

10.45 hydrated sodium carbonate(Na₂CO₃.xH₂O) were heated until 3.87 of 3.87of anhydrous (Na₂CO₃) remained, so

mass H₂O released :

\tt 10.45-3.87=6.58~g

mass Na₂CO₃ = 3.87 g

mol ratio Na₂CO₃(MW= g/mol) : H₂O(MW=18 g/mol) =

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6 0
2 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil
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Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

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We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

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Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

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