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!

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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