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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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Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and
user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity  of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2  °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J  = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0
2 years ago
An ice cube measuring 5.80 cm by 5.80 cm by 5.80 cm has a density of 0.917 g/mL What is the mass?
OLga [1]

Answer: 178.9 g

Explanation:

Density = \frac{mass}{volume in mL}

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

0.917 g/mL = (mass) / (195.112 mL)

and solve for mass!

3 0
1 year ago
A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa
bogdanovich [222]

Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

6 0
2 years ago
The hybrid orbital set used by the central atom in h2co is ________. the hybrid orbital set used by the central atom in h2co is
Feliz [49]
The Lewis structure for H₂CO is shown in the attached picture. The central atom is the carbon. However, I'm not sure which bond you're referring to. There can be two answers. The two C-H bonds are sp³ hybridized because it is a single bond. The C=O bond is sp² hybridized because it is a double bond.

5 0
2 years ago
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
2 years ago
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