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serious [3.7K]
3 years ago
11

If you pump 40.088g of octane (C8H18) into your gas tank, how mant molecules of octane are you pumping

Chemistry
2 answers:
Sonbull [250]3 years ago
6 0

Answer:

2.11 × 10²³ molecules of octane

Explanation:

Given data:

Mass of octane = 40.088 g

Molecules of octane = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 40.088 g / 114.23 g/mol

Number of moles =  0.351 mol

Number of molecules:

one mole of octane = 6.022 × 10²³ molecules of octane

0.351 mole of octane = 0.351 ×  6.022 × 10²³ molecules of octane

2.11 × 10²³ molecules of octane

olga_2 [115]3 years ago
6 0

2.11 x 10²³molecules

Explanation:

Given parameters:

Mass of octane = 40.088g

Unknown:

Number of molecules of the octane = ?

Solution:

To find the number of molecules of the octane. We need to first convert the given mass of octane to number of moles.

  Number of moles = \frac{mass}{molar mass}

Molar mass of octane = (8 x 12)  + (1 x 18) = 114g/mol

 Number of moles = \frac{40.088}{114} = 0.35moles

1 mole of a substance contains  6.02 x 10²³ molecules

0.35 moles will contain 0.35 x 6.02 x 10²³ = 2.11 x 10²³molecules

Learn more:

Number of molecules brainly.com/question/10818009

#learnwithBrainly

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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

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Therefore, the mole fraction of Xe is 0.67.

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Your theoretical yield is 81.2 grams, and your actual yield is 78.2 grams. What is the percent yield?
ioda

Answer:

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Explanation:

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