1)Fluorine-21 has a half life of approximately 5 seconds. If you start with 100g, how much would remain after 10 seconds?
What’s the answr

25 g

Explanation:

The following data were obtained from the question:

Half life (t½) = 5 s

Original amount (N₀) = 100 g

Time (t) = 10 s

Amount remaining (N) =..?

Next, we shall determine the decay constant for the disintegration. This can be obtained as follow:

Half life (t½) = 5 s

Decay constant (K) =.?

K = 0.693 / t½

K = 0.693 / 5

K = 0.1386 s¯¹

Finally, we shall determine the amount of the isotope that remained after 10 s as follow:

Original amount (N₀) = 100 g

Time (t) = 10 s

Decay (K) = 0.1386 s¯¹

Amount remaining (N) =..?

Log (N₀/N) = kt /2.303

Log (100/N) = (0.1386 × 10) / 2.303

Log (100/N) = 1.386 / 2.303

Log (100/N) = 0.6018

Take the anti log of 0.6018

100/N = antilog (0.6018)

100/N = 4

Cross multiply

100 = N × 4

Divide both side by 4

N = 100/4

N = 25 g

Therefore, the amount remaining after 10 s is 25 g

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.