The half life of a radioactive element is 8 hours. A sample of the element is tested and found to contain 6g of the element. How much of the element was present 32 hours before the element was tested? 0.375g 6g 48g 96g

m₀ = 96 g

Explanation:

First, we will calculate the no. of half-lives passed:

where,

n = no. of half-lives passed = ?

T = Total time elapsed = 32 h

= Half-Life = 8 h

Therefore,

Now, for the initial amount of element:

where,

m₀ = initial amount of element = ?

m = current amount of element = 6 g

Therefore,

m₀ = (6 g)(16)

m₀ = 96 g

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .