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.
A. The breeze blow from higher pressure over the water to the lower
Answer:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Explanation:
Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:
3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O
Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal
Answer:
Los usos de la plata son cientos, sobre todo en procesos industriales, comerciales y hasta personales. Su resistencia a la corrosión la hace ideal para la elaboración de recipientes especiales o para recubrir otros metales.
Explanation:
Answer:
Ag 0 is the reducing agent.
Explanation:
Reducing -> gaining electrons
Oxidizing -> losing electrons
Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.
