Krypton-79 has a half-life of 35 hours. how many half-lives have passed after 105 hours?
1 answer:
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Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+
⇄
Formation constant Kf
Kf = / ( [
][
] ) = 5.0 × 10¹⁰
Now,
[] =
; ∝₄ = 0.35
so the equilibrium is;
+
⇄
+ 4H⁺
Given that; =
{ 1 mol
reacts with 1 mol
}
so at equilibrium, =
= x
∴
+
⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [
][
] ) =
/ ( [
][
] ) = 5.0 × 10¹⁰
⇒ / ( [
][
] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Answer:
0.88g
Explanation:
The reaction equation:
2NaI + Cl₂ → 2NaCl + I₂
Given parameters:
Mass of Sodium iodide = 2.29g
Unknown:
Mass of NaCl = ?
Solution:
To solve this problem, we work from the known to the unknown.
First find the number of NaI from the mass given;
Number of moles =
Molar mass of NaI = 23 + 126.9 = 149.9g/mol
Now insert the parameters and solve;
Number of moles = = 0.015mol
So;
From the balanced reaction equation;
2 moles of NaI produced 2 moles of NaCl
0.015mole of NaI will produce 0.015mole of NaCl
Therefore;
Mass = number of moles x molar mass
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Now;
Mass of NaCl = 0.015 x 58.5 = 0.88g