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fredd [130]
1 year 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]1 year 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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sertanlavr [38]

Answer:

Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.

hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond  must be oriented in  the opposite directions with respect to each other.]

So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.

If one isomer of the alkene is trans then the other two isomers may be cis .

Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.

The two possibility of cis structures are possible:

in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.

Or the other way could be that two  chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.

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7 0
2 years ago
Give the number of significant figures in this number: 40.00
Tatiana [17]

A significant figure is every symbol that made the number itself.

In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.

If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

4 0
2 years ago
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The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will
yulyashka [42]

C) change to water at the same temperature

Explanation:

Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.

  • The heat of fusion is the amount of energy needed to melt a given mass of a solid
  • It is also conversely the amount of energy removed from a substance to freeze it.
  • The addition of this energy does not cause a decrease or increase in temperature.
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Learn more:

Heat of fusion brainly.com/question/4050938

#learnwithBrainly

3 0
2 years ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
1 year ago
Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0
2 years ago
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