1. The half-life of Co-60 is 5.3 years. If the initial mass of the sample is 1000g, how much will remain after 15.9 years.

125 g

Explanation:

The following data were obtained from the question:

Half-life (t½) = 5.3 years.

Original amount (N₀) = 1000 g

Time (t) = 15.9 years.

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 5.3 years.

Time (t) = 15.9 years.

Number of half-lives (n) =?

n = t / t½

n = 15.9 / 5.3

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the amount of Co–60 that is remaining. This can be obtained as follow:

Original amount (N₀) = 1000 g

Number of half-lives (n) = 3

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2³ × 1000

N = 1/8 × 1000

N = 0.125 × 1000

N = 125 g

Therefore, the amount of Co–60 that is remaining is 125 g

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.