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.
1, 2, and 3
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Answer:
Objects appear different colours because they absorb some colours (wavelengths) and reflected or transmit other colours. ... For example, a red shirt looks red because the dye molecules in the fabric have absorbed the wavelengths of light from the violet/blue end of the spectrum
Al(NO3)3 + 3KOH -------> 3KNO3 + Al(OH)3
50 ml * .2 moles/ liter = .01 Moles of Al(NO3)3
200 ml * .1 moles/liter = .02 Moles of KOH
Since the ratio between the two reactants according to the chemical equation is 1:3, we would need .03 moles of one to fully react with .01 moles of the other. Since we don't, only 1/150 mole of the first reactant will react with the .02 moles of the second reactant. This will produce .02 moles of KNO3 as well as .01 moles of Al(OH)3
.02 moles KNO3 = .02(48 grams + 14 grams + 40 grams) = .02(102 grams) = 2.04 grams
