To solve this kinematics formula use the following equation:
Vf = Vi + at
Vf = 0 + (9.81 m/s^2)(3 seconds)
Vf = 29.43 m/s and or about 29.4 m/s of reported to 3 significant figures.
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.
The answers are low concentrated (dilute) and high concentrated respectively.
As the low concentrated salt solution has a higher water potential than that of the high concentrated salt solution, water molecules will flow from the region of higher water potential to the region of lower water potential, thus from the dilute salt solution to the high concentrated salt solution. This is due to the movement called osmosis. Note that osmosis also requires water to flow through a differentially permeable membrane, which means the membrane can allow certain substances (not all) to go in or out. If the differentially permeable membrane is not present, the movement of water molecules may be regarded as diffusion.
Therefore, the answers for the blanks are low concentrated and high concentrated.
Answer:
2.15 L
Explanation:
M(NaOCl) = 23.0 + 16.0 + 35.5 = 74.5 g/mol
120.3 g *1 mol/74.5 g = 1.615 mol NaOCl
0.750 M = 0.750 mol/L
1.615 mol * 1L/0.750 mol = 2.15 L
