# 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.)

### 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

## Related Questions

The periodic table displaysOA. all of the known elements that exist in the world today.
OB. only the important elements that exist in the world.
OC. only the important compounds that exist in the world.

OA. all the known elements that exist in the world today.

At this location, it is summer in the Northern Hemisphere because the North Pole is tiltedAway from the sun

Towards the sun

This happens twice a year during Earth's orbit. Near June 21 the north pole is tilted 23.5 degrees toward our Sun and the northern hemisphere experiences summer solstice, the longest day of the northern hemisphere year.

...

Do other planets have seasons?

Uranus

30,589

97.8

Spring Equinox* 2050

Summer Solstice*

points toward the sun.

closer the earth is to the sun the more  hot it will be the closer it is to summer,

you can see at D northern hemisphere is closest to sun and the north pole is pointing toward the sun.

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?

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

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

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

B

Explanation:

The Agricultural Revolution gave Britain the most productive agriculture in Europe, with 19th-century yields as much as 80% higher than the Continental average. ... By the 19th century, marketing was nationwide and the vast majority of agricultural production was for the market rather than for the farmer and his family.

hope this helps :)

What is the hydrogen-ion concentration of the ph is 3.7

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

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.

Heat gained by ice

heat gained in melting + heat gained in getting warmed

= mass x latent heat + mass x specific heat x rise in temperature

= 17.69 x 80 + 17.69 x 1 x ( 12.9 - 0 )

= 1643.4 Cals

Heat lost by water

=  mass x specific heat x fall in temperature

98.67 x 1 x ( 28.77 - 12.9 )

= 1565.89 Cals

Heat lost by calorimeter

heat capacity x fall in temperature

X x ( 28.77 - 12.9 )

Heat gained = heat lost

1643.4 = 1565.89 +15.87X

X = 4.88 Cals per degree celsius