...because this notation increases the convenience in using the numbers
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.
Answer:
It's Al 3+ + 3e - → Al
Explanation:
Because three negatively charged electrons are needed to balance the three positive charges on the aluminum ion
Answer:
Explanation:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol= e-
Mass= 9.10938356×10-31 Kg
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
He constructed the glass tube and create vacuum in it. He applied electric current between electrodes. He noticed that a ray of particles coming from cathode to wards positively charged anode. This ray was cathode ray.
Properties of cathode ray:
The ray is travel in straight line.
The cathode ray is independent of composition of cathode.
When electric field is applied cathode ray is deflected towards the positively charged plate.
Hence it was consist of negatively charged particles.
Answer:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>