Maybe like a diamond or something
1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u
Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.
1) We can find the next equation:
mass=molar mass x number of moles.
x=number of moles of helium
y=number of moles of helium.
(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:
(1)
4x+83.8y=103.75
2) We can find other equation:
We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.
1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).
Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.
x=30% of (x+y)
Therefore, we have this other equation:
(2)
x=0.3(x+y)
With the equations(1) and (2), we have the next system of equations:
4x+83.8y=103.75
x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7
We solve this system of equations by substitution method.
x=3y/7
4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21
x=3y/7=3(1.21)/7=0.52
We have 0.52 moles of helium and 1.21 moles of Krypton.
1 mol=6.022 * 10²³ atoms
Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).
Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.
Answer: The sample have 1.044 * 10²⁴ atoms.
I think the answer is option 2
Answer:
<span><span>22688</span>Ra→<span>l<span>22286</span></span>Rn+<span>l<span>42</span></span>He</span>
Explanation:
An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.
During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.
Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation:
<span><span>22688</span>Ra→<span>l<span>22286</span></span>Rn+<span>l<span>42</span></span>He</span>
Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.
Also, the sum of the superscripts (masses) is the same on each side of the equation.
Now try figuring out Am-241
