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:
ΔT = 20.06 °C
Explanation:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 1.17 kJ = 1170 J
m = mass = 24.1 g
Cp = Specific Heat Capacity = 2.42 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = <u>??</u>
Solving eq. 1 for ΔT,
ΔT = Q / m Cp
Putting values,
ΔT = 1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹
ΔT = 20.06 °C
Answer:
P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm
Explanation:
Apply Dalton's Law of Partial Pressures.
P(total) = ∑Partial Pressures = ∑(256mm + 198mm + 48mm) = 504 mmHg
P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm
