Answer:
B
Explanation:
hợp chất nào có khả năng là hợp chất ion cao nhất ?
A Al2O3?
B
Molarity is calculated by using following formula,
Molarity = Moles / Volume
Data Given:
Moles = 23 moles
Volume = 100 ml ÷ 1000 = 0.1 L
Putting values in eq. 1,
Molarity = 23 mol / 0.1 L
Molarity = 230 mol/dm³
Result:
When 23 mol of solute is dissolved in a solvent to make a solution of 100 ml, then it will have a Molarity of 230 mol/dm³.
12.5% of strontium-90 would remain in a sample after three half-lives have passed. Half-life automatically means 50% of the original amount would remain.
Answser:
3.77 mg of K-40 decayed into Ar-40.
Data:
1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90%<span> of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.</span>
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg
Soltuion:
1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed
=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg
=> x = 0.377 mg / 0.1 = 3.77 mg
That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.
Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.
2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>
mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg
3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg
4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg
5) amount of K-40 that decayed into Ar-40 as percent
% = [3.77 mg / 4.58mg] * 100 = 82.31 %.
