If 800g of a radioactive substance are present initially and 8 years later only 450g remain, how much of the substance will be present after 16 years? (Round answer to a whole number)

A=Pe^(rt)

P = 800g

t = 8 years

A = 450g

r = This is what we will try and find to start with

450=800e^(r*8)

After running the math through a calculator, we end with r = -0.07192

Now we just re-input this information into our equation: A=800e^(-0.07192*16)

A=800e^(1.15072)

Now we will re-write the equation using the negative exponent rule:

A = 800 1/e^1.15072

Combine right side:

A = 800/e^1.15072

Then do the math:

A = 253.12709836

That will give us A = 253 (rounded to the whole number)

I hope this helps! :)

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

x coordinate is none

step-by-step explanation: