Question

The combustion of a single molecule of methane produces about 10 ev of energy. a methane molecule has a mass of 16 amu. the fission of a uranium-235 atom produces about 189 mev of energy. based on these numbers, about what mass of methane will produce as much energy as a single gram of uranium-235?

Answer 1

About 0.05275 grams of methane will produce the same amount of energy as a single gram of uranium-235.

Step-by-step explanation:

The question asks us to determine the mass of methane that will produce the same amount of energy as a single gram of uranium-235. We are given that the combustion of a single methane molecule produces about 10 ev of energy and that a methane molecule has a mass of 16 amu. On the other hand, the fission of a uranium-235 atom produces about 189 mev of energy. To find the mass of methane that will produce the same amount of energy as a gram of uranium-235, we can set up the following proportion:

(mass of methane)/(mass of uranium-235) = (energy produced by methane)/(energy produced by uranium-235)

Using the given values:

(mass of methane)/(1 g) = (10 ev)/(189 mev)

Simplifying and converting the energy values to the same units:

(mass of methane)/(1 g) = (10 × 6.242 × 10^18 eV)/(189 × 6.242 × 10^18 eV)

Calculating the values:

• 10 × 6.242 × 10^18 = 62.42 × 10^18 eV
• 189 × 6.242 × 10^18 = 1182.738 × 10^18 eV

Now we can plug in the values into the proportion:

(mass of methane)/(1 g) = (62.42 × 10^18)/(1182.738 × 10^18)

Simplifying:

mass of methane = (1 g) × (62.42 × 10^18)/(1182.738 × 10^18)

mass of methane = 0.05275 g

Therefore, about 0.05275 grams of methane will produce the same amount of energy as a single gram of uranium-235.

Answer 2

If one molecule of methane produces 10eV then we need:

`n=(189MeV)/(10eV)`,
molecules of methane to get 189MeV.

`n=(189MeV)/(10eV)=(189)/(10)\cdot 10^6=18900000`
To get the mass we simply multiply this number by the mass of one molecule of methane.

`m=16\cdot18900000=302400000amu`
One atomic mass unit is:

`amu=1.66053904\cdot 10^(-24)$g`
There we would need:

`m=302400000\cdot1.66053904\cdot 10^(-24)=5.0214701\cdot10^(-16)$g`
grams of methane.