1070 hours.
1 mole of iron-59 would mass 59 grams, so 0.133 picograms would be 0.133x10^-12 / 59 = 2.25x10^-15 moles of iron-59. Multiplying by Avogadro's number, we can determine the number of atoms of iron-59 we have, so: 2.25x10^-15 * 6.02214x10^23 = 1.35x10^9
Since we have 242 decays over a period of 1 second, we can divide the
number of atoms left by the original number of atoms
(1350000000 - 242)/1350000000
= 1349999758/1350000000
= 0.999999820740741
And calculate the logarithm to base 2 of that quotient.
ln(0.999999820740741)/ln(2)
= -1.79259275281191x10^-7/0.693147180559945
= -2.58616467481524x10^-7
The reciprocal of this number will be the half life in seconds. So
-1/2.58616467481524x10^-7
= -3866729.79388461
And dividing by 3600 (number of seconds in an hour) will give the half-life in
hours.
-3866729.79388461 / 3600 = -1074.091609
So the half life in hours to 3 significant figures is 1070 hours.
Dividing that figure by 24 gives a half life of 44.58 days which is in pretty close agreement to the official half-life of 44.495 days for iron-59.
Answer: Hello, There! Your Answer is Below
2.4 x 1024 ions
Explanation:
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g
X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol
X = 2.38 × 10²⁴ Cl⁻ Ions
Hope this Helps!
Have a great DAy!
`August~
The ionic eqn is as follow:
1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)
3moles of No3- ion on both sides cancels out to give the net ionic eqn:
1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)
