D. Two electrons in its first energy level; eight electrons in its second energy level; six valence electrons in its outermost energy level.
Please correct me if I'm wrong!! :)
M(H2SO4)=n*M=4.75*98=465.5g
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.
D:
When electrons are gained, the charge of the atom decreases.
When you are given an atom with a charge, the oxidation of that atom is the charge. So by going from a Cr^3+ (Oxidation Number = 3) to a Cr^2+ (Oxidation Number = 2), the Oxidation Number thus decreases.
