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Vsevolod [243]
4 years ago
15

Calculate the number of pounds of CO2CO2 released into the atmosphere when a 22.0 gallon22.0 gallon tank of gasoline is burned i

n an automobile engine. Assume that gasoline is primarily octane, C8H18,C8H18, and that the density of gasoline is 0.692 g⋅mL−1.0.692 g⋅mL−1. This assumption ignores additives. Also, assume complete combustion. CO2CO2 released:
Chemistry
1 answer:
Gnesinka [82]4 years ago
3 0

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

Density= \frac{Mass}{Volume}

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

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

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane =\frac{57,629.081 g}{114.08g/mol}=505.1637 mol

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

\frac{16}{2}\times 505.1637 mol=4,041.3100 mol of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

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3 years ago
SHOW WORK AND INCLUDE UNITS
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Answer:

36365.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C), and change in temperature (Φ)

Thus, Q = MCΦ

Since, M = 45.4 g

C = 3.56 J/g°C,

Φ = 250°C - 25°C = 225°C

Q = 45.4g x 3.56J/g°C x 225°C

Q= 36365.4 Joules

Thus, 36365.4 Joules of heat energy is released when the lithium is cooled.

6 0
3 years ago
The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are
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Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

                                          [A] = - Kt + [Ao]

where:

  • [A]: concentration of the reactant A at the <em>t </em>time,
  • [A]o: initial concentration of the reactant A,
  • K: rate constant,
  • t: elapsed time of the reaction

<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

                                        t  = ([Ao] - [A])/K

<em>We replace the numerical values:</em>

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0
3 years ago
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