Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 6.0 days and 2 hours, how many milligrams of thallium-201 will remain?

Explanation:

Half-life = 73 hours

Disintegration period = 6 days 2 hours = (6*24) hours + 2 hours

= 144 hours + 2 hours

= 146 hours

No. of Half lives = 146 hours / 73 hours

No. of Half lives = 2

Hence, there are 2 half lives passed in 6 days 2 hours.

Amount of thallium-201 left after first half-life:

= 4.0 mg / 2

= 2.0 mg

Amount of thallium-201 left after second half-life:

= 2.0 mg / 2

= 1.0 mg

Hence, the amount of thallium-201 left after 6.0 days and 2 hours is 1.0 mg.

Hope this helped!

its mg hope this

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g