Respuesta :
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.
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.
Answer: The total number of particles present in the sample are [tex]53.815\times 10^{23}[/tex] particles.
Explanation:
We are given a total mass of the sample 103.75 grams in which 30% are helium atoms and 70% are krypton atoms.
The mass of 30% of helium atoms in the given amount of sample is = [tex]\frac{30}{100}\times 103.75=31.125g[/tex]
the mass of 7% of krypton atoms in the given amount of sample is = [tex]\frac{70}{100}\times 103.75=72.625g[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] atoms/particles.
- For Helium:
1 mole of helium contains 4 grams.
4 grams of helium element contain [tex]6.022\times 10^{23}[/tex] atoms/particles.
So, 31.125 g of helium element contain [tex]\frac{6.022\times 10^{23}}{4}\times 31.125=46.858\times 10^{23}[/tex] atoms/particles.
- For Krypton:
1 mole of krypton contains 84 grams.
84 grams of krypton element contain [tex]6.022\times 10^{23}[/tex] atoms/particles.
So, 72.625 g of krypton element contain [tex]\frac{6.022\times 10^{23}}{84}\times 72.625=5.026\times 10^{23}[/tex] atoms/particles.
Total number of particles in the given sample = [tex](46.858+5.026)\times 10^{23}=51.884\times 10^{23}[/tex] atoms/particles.
