Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?
Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.
V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3
Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":
9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)
= 6.96 x 10^19 atoms of Ir-122 are present.
Option no. 2 subtracting the atomic number from the mass number
Example: Zn - 65 atomic weight or mass/mass number
30 atomic number
65-30 = 35 number of neutrons
Answer:
526.85K
Explanation:
Based on Charles's law, the volume of a gas is directly proportional to absolute temperature. The formrula is:
V₁ / T₁ = V₂ / T₂
<em>Where 1 represents the initial state and 2 the final state of the gas</em>
Using the values of the problem:
V₁ = 10.0L
T₁ = 127°C + 273.15K = 400.15K
V₂ = 20.0L
Thus, replacing in the formula:
10.0L / 400.15K = 20.0L / T₂
T₂ = 800K
In Celsius:
800K - 273.15 =<em> 526.85K</em>
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Answer:
2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.
Explanation:
The reaction:
<em>2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.</em>
2.0 moles of hydrogen gas react with 1.0 mole of oxygen gas to produce 2.0 moles of water vapor in presence of Pt as a catalyst.
Question:
1. (NH)2CrO
a) Number of moles of H:
b) Number of moles of N:
Answer:
a) Number of moles of H: 2
b) Number of moles of N: 2
Explanation:
The 
is ammonium Chromate which is monoclinic and yellow Crystal that is formed due to the reaction of ammonium Hydroxide and ammonium di-chromate. It is used as catalyst, corrosion inhibitor as well as analytical inhibitors.
Question:
2. Ag.SO.
a) Molar Mass:
b) Percent Composition of Ag:
c) Percent Composition of S:
d) Percent Composition of O:
Answer:
a) Molar Mass: 155.93 Kg
b) Percent Composition of Ag: 69%
c) Percent Composition of S: 20.5%
d) Percent Composition of O: 10.2%
Explanation:
Molar mass = molar mass of Ag + molar mass of S + molar mass of O
=>107.87+32.06+16
=> 155.93 Kg
Percent Composition of Ag
= 
= 
= 0.69 \times 100
= 69%
Percent Composition of S:
= 
=
= 0.205 \times 100
= 20.5%
Percent Composition of O:
= 
= 
= 0.102 \times 100
= 10.2%
