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SVETLANKA909090 [29]
3 years ago
10

) A vehicle owner fuels his/her car with pure octane. What quantity of CO2 (in kg) goes into our atmosphere in one year of their

automobile use? (This person spends $1500 per year on gasoline, and that gas costs $3.00 per gallon) Show your work. 1 gallon of gas weighs 6 pounds. 1 kg is 2.2 pounds.
Chemistry
2 answers:
vodomira [7]3 years ago
7 0
I’ll send someone to come and help you
jekas [21]3 years ago
6 0

Money spent on fuel in 1 year is $1500, the cost of fuel is $3.00 per gallon. Thus, gallon of fuel used in 1 year can be calculated as follows:

$3 =1 gallon

$1=\frac{1}{3}gallons

Thus, $1500=\frac{1500}{3}gallons=500 gallons

converting this into m^{3}

1 gallon=0.0037 m^{3}

Thus,

500 gallons=1.8927 m^{3}

The fuel used is pure octane, the density of octane is 703 kg/m^{3} thus, mass of octane used can be calculated as follows:

m=d×V=703 kg/m^{3}×1.8927 m^{3}=1.33\times 10^{3} kg or 1.33\times 10^{6} g

Now, molar mass of octane is 114.23 g/mol

converting mass into number of moles as follows:

n=\frac{m}{M}=\frac{1.33\times 10^{6}g}{114.23 g/mol}=1.167\times 10^{4}mol

The reaction for combustion of octane is as follows:

2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O

From the balanced reaction, 2 moles of octane gives 16 moles of CO_{2} thus, 1 mole of octane will give \frac{16}{2}=8 moles of CO_{2}.

1.167\times 10^{4}mol of octane =8\times 1.167\times 10^{4}mol=9.34\times 10^{4} mol of CO_{2}.

Now, molar mass of CO_{2} is 44.01 g/mol, mass can be calculated as follows:

m=n×M=9.34\times 10^{4} mol ×44.01 g/mol=4.11×10^{6} g.

converting into kg, mass of CO_{2} produced will be 4.11×10^{3} kg.

Thus, quantity of CO_{2} goes into our atmosphere in 1 year is 4.11×10^{3} kg.

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how to determine the net charge of the tripeptide Asp-Gly-Leu at pH 7. Can someone show in details and tricks on how to solve it
Ugo [173]

Answer:

0!

Explanation:

  • You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:            

           

                   -COOH         -NH3              

pH 7------------------------------------------------------              

Asn               -                      +

Gly                -                      +

Leu               -                      +

  • Now that we have our table we'll sketch our peptide's structure:

<em>HN-Asn-Gly-Leu-COOH</em>

This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:

+1 (HN-Asn)

-1 (Leu-COOH)

=0

The net charge is 0!

I hope you find this information useful and interesting! Good luck!

5 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.
kykrilka [37]

Answer-The correct option is option d with says all of the above.

Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.

The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.

5 0
3 years ago
identify the part of the slow carbon cycle in whitch the total amount of carbon is most likely decreasing and explain why this d
Westkost [7]

Answer:

i just had gthe sme question looked it up and got it WRONG

Explanation:

sorry i ont have a clue

3 0
3 years ago
Which sample has particles with the lowest average kinetic energy?
Anastasy [175]
The question asks average kinetic energy. So it is only related with the temperature. The higher temperature is, the higher kinetic energy is. So the answer is (4).
3 0
3 years ago
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