A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Answer 1
Answer:

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

Learn more about Half-life at: brainly.com/question/4702752


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Answers

Answer:

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Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing that the graviatational potential energy of falling water per second = mass (kg) x height (m) x gravity (9.8 m/s2), what is the power of Niagara Falls? How many 15 W LED light bulbs could it power?

Answers

Answer:

1. 176 × 10^12 W ; 78400000000

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Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

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Answers

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Answers

Answer:

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Explanation:

One way to experimentally measure the heat capacity of a Styrofoam cup calorimeter would be to melt a known mass of ice in warm water and measure hte temperature change. Use the data below to determine the experimental heat capacity of the calorimeter. Use the literature heat of fusion for ice in your calculations. Assume the ice added is at 0.00 c.Mass of ice added: 17.69gMass of water in calorimeter: 98.67gT-Initial of water: 28.7T-Final of water after melting ice: 12.9C

Answers

Answer:

4.88 Cals per degree celsius

Explanation:

We have taken heat of fusion of ice = 80 cals / g

We have taken speciic heat of water = 1 cal/g per degree celsius

In this experiment ,  let the heat capacity of calorimeter be X.

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Heat lost by water

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Heat lost by calorimeter

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