An original sample of K-40 has a mass of 25.00 grams.
After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged.
What is the half-life of K-40?
First step is to determine the remaining decimal amount.
3.125 grams /25.00 grams = 0.125
Second step is to determine the number of half lives.
(1/2)^n = 0.125
N log (1/2) = log 0.125
N = 3 years
Answer is: <span> two samples have in common same amount of substance and same number of particles.
1) There are same amount of substance in both beakers:
n(Zn) = 1 mol.
n(ZnCl</span>₂) = 1 mol.
2) There are same number of particles (atoms, molecules, ions) in both beakers:
N(Zn) = n(Zn) · Na.
N(Zn) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ atoms of zinc.
N(ZnCl₂) = n(ZnCl₂) · Na.
N(ZnCl₂) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ molecules of zinc(II) chloride.
Na - Avogadro number.
If the sperm carries a Y chromosome, it will result in a male. During fertilization, gametes from the sperm combine with gametes from the egg to form a zygote. The zygote contains two sets of 23 chromosomes, for the required 46.
Because neither can be created or destroyed
