Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 5.0 °, how many milligrams of th
1 answer:
<u>Given:</u>
Half life of Thallium-201 (T1/2)= 73 hrs
Initial amount of thallium (A₀) = 4.0 mg
Disintegration Time period (t) = 5.0 hrs
<u>To determine:</u>
Amount of thallium remaining, A
<u>Explanation:</u>
The radioactive half-life (T1/2) can be related to the amounts of the radio active element through the relation:
A =
A = 4.0 * = 3.815 mg
Ans: Amount of thallium-201 remaining is 3.815 mg
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Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
10 ml 17.50 ml
(x) M 0.200 M
Molarity =
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
=
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration =
Molar Concentration =
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M