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Maksim231197 [3]
3 years ago
11

Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many

grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Chemistry
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

Explanation:

Number of moles of propane:

=Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

= 44 * 6.8181

= 299.9964 grams

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If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh
RideAnS [48]

Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve  2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})

Where Cs is the concentration in the surface of the solid

Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055

Then we solve for x:

x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm

6 0
3 years ago
imagine you are a particle of air explain your journey in the atmosphere as you beacome a heated particle if you answer it clear
enyata [817]
Well when a particle of air is becomes heated it rises, right? So you could write some like you started off close to the earth (aka the troposphere) until you became heated then you started to rise and as you reached higher elevations you cooled down and you were recycled into cool air and you moved back down and became new fresh cool air until the next time you'll become heated and rise again to be recycled into fresh cool new air.
6 0
3 years ago
Imagine that A and B are cations and X, Y, and Z are anions, and that the following reactions occur:
Gekata [30.6K]
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
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AX(aq) is soluble and <span>BY(aq) is insoluble

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5 0
4 years ago
When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o
Angelina_Jolie [31]

19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.

Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2

Volume of O_2 (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of O_2 (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2C_4H_10 + 13O_2 ---> 8CO_2 + 10H_2O

From reaction;

13 moles O_2 require 2 moles C_4H_10

So, 1.663 moles O_2 will require = 2 x 1.663 ÷13 = 0.256 moles of C_4H_10

Thus C_4H_10 is a limiting reagent. So it will drive the yield of CO_2.

Moles of CO_2 produced = (8/2) 0.2 = 0.8 moles

Pressure of CO_2 (P) = 102 - 2.24 = 99.76 kPa = 99.76  ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for CO_2,

PV = nRT

0.985 V = 0.8 0.0821 x 295

V = 19.7 liter

The volume of CO_2 produced = 19.7 liter.

Learn more about the vapour pressure here:

brainly.com/question/25699778

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5 0
2 years ago
Be sure to answer all parts. One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, whic
KIM [24]

Answer:

The answer is below

Explanation:

one gallon of gasoline produces 9.50 kg of carbon.

The total number of cars = 40 million

Distance covered by each car = 7930 miles

Consumption rate of the cars per miles traveled is 23.6 miles per gallon.

Hence the annual gasoline consumption by all the cars in the United States of America = (total number of cars × Distance covered by each car) ÷ Consumption rate of the cars per miles

annual gasoline consumption by all the cars = (40000000 × 7930 miles) ÷ 23.6 miles/gallon = `1.344067797 × 10¹⁰ gallons

1.344067797 × 10¹⁰ gallons = 1.344067797*10^{10}\  gallons *\frac{9.50\ kg}{1\ gallon} =1.276864407*10^{11}\ kg\ of \ carbon\ dioxide

4 0
3 years ago
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