Answer:
Explanation:
91.4
grams
Explanation:
C
=
m
o
l
v
o
l
u
m
e
2.45
M
=
m
o
l
0.5
L
2.45
M
⋅
0.5
L
=
m
o
l
m
o
l
=
1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225
m
o
l
⋅
(
39.1
+
35.5
)
g
m
o
l
1.225
m
o
l
⋅
74.6
g
m
o
l
=
1.225
⋅
74.6
g
=
91.4
g
In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. ... Electrical forces are extremely sensitive to distance.
Soil covered, saturated, submerged, flooded w water, standing water
The answer is a , as aluminium has 13 protons and electrons .
The partial pressure (Px) of a gas in a gas mixture is equal to its mole fraction (Xi) multiplied by the total pressure (P) of the gas mixture. That means that we have to calculate the mole fraction of each gas, then calculate its partial pressure. The mole fraction of a gas is its number of moles (n) divided by the total number of moles.
![$$ Mole fraction of methane: \\$\chi_{\text {methane }}=\frac{\mathrm{n}_{\text {methane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.24 \mathrm{~mol}+0.421 \mathrm{~mol}+0.116 \mathrm{~mol}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {methane }}=0.938$](https://tex.z-dn.net/?f=%24%24%20Mole%20fraction%20of%20methane%3A%20%5C%5C%24%5Cchi_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D%5Cfrac%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Bmethane%20%7D%7D%7D%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Btotal%20%7D%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D%5Cfrac%7B8.24%20%5Cmathrm%7B~mol%7D%7D%7B8.24%20%5Cmathrm%7B~mol%7D%2B0.421%20%5Cmathrm%7B~mol%7D%2B0.116%20%5Cmathrm%7B~mol%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D%5Cfrac%7B8.24%20%5Cmathrm%7B~mol%7D%7D%7B8.78%20%5Cmathrm%7B~mol%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D0.938%24)
![$$Partial Pressure of methane:\\$\mathrm{P}_{\text {methane }}=\chi_{\text {methane }} \times \mathrm{PP}_{\text {methane }}=0.938 \times 1.37 \mathrm{~atm} \mathbf{P}_{\text {methane }}=\mathbf{1 . 2 8} \mathbf{~ a t m}$](https://tex.z-dn.net/?f=%24%24Partial%20Pressure%20of%20methane%3A%5C%5C%24%5Cmathrm%7BP%7D_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D%5Cchi_%7B%5Ctext%20%7Bmethane%20%7D%7D%20%5Ctimes%20%5Cmathrm%7BPP%7D_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D0.938%20%5Ctimes%201.37%20%5Cmathrm%7B~atm%7D%20%5Cmathbf%7BP%7D_%7B%5Ctext%20%7Bmethane%20%7D%7D%3D%5Cmathbf%7B1%20.%202%208%7D%20%5Cmathbf%7B~%20a%20t%20m%7D%24)
![$$Mole fraction of ethane: \\$\chi_{\text {ethane }}=\frac{\mathrm{n}_{\text {ethane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {ethane }}=\frac{0.421 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {ethane }}=0.0479$](https://tex.z-dn.net/?f=%24%24Mole%20fraction%20of%20ethane%3A%20%5C%5C%24%5Cchi_%7B%5Ctext%20%7Bethane%20%7D%7D%3D%5Cfrac%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Bethane%20%7D%7D%7D%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Btotal%20%7D%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bethane%20%7D%7D%3D%5Cfrac%7B0.421%20%5Cmathrm%7B~mol%7D%7D%7B8.78%20%5Cmathrm%7B~mol%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bethane%20%7D%7D%3D0.0479%24)
![$$Partial pressure of ethane:\\$\mathrm{P}_{\text {ethane }}=\chi_{\text {ethane }} \times \mathrm{PP}_{\text {ethane }}=0.0479 \times 1.37 \mathrm{~atm} \mathrm{P}_{\text {ethane }}=\mathbf{0 . 0 6 5 6} \mathbf{~ a t m}$](https://tex.z-dn.net/?f=%24%24Partial%20pressure%20of%20ethane%3A%5C%5C%24%5Cmathrm%7BP%7D_%7B%5Ctext%20%7Bethane%20%7D%7D%3D%5Cchi_%7B%5Ctext%20%7Bethane%20%7D%7D%20%5Ctimes%20%5Cmathrm%7BPP%7D_%7B%5Ctext%20%7Bethane%20%7D%7D%3D0.0479%20%5Ctimes%201.37%20%5Cmathrm%7B~atm%7D%20%5Cmathrm%7BP%7D_%7B%5Ctext%20%7Bethane%20%7D%7D%3D%5Cmathbf%7B0%20.%200%206%205%206%7D%20%5Cmathbf%7B~%20a%20t%20m%7D%24)
![$$Mole fraction of propane:\\$\chi_{\text {propane }}=\frac{\mathrm{n}_{\text {propane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {propane }}=\frac{0.116 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {propane }}=0.0132$](https://tex.z-dn.net/?f=%24%24Mole%20fraction%20of%20propane%3A%5C%5C%24%5Cchi_%7B%5Ctext%20%7Bpropane%20%7D%7D%3D%5Cfrac%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Bpropane%20%7D%7D%7D%7B%5Cmathrm%7Bn%7D_%7B%5Ctext%20%7Btotal%20%7D%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bpropane%20%7D%7D%3D%5Cfrac%7B0.116%20%5Cmathrm%7B~mol%7D%7D%7B8.78%20%5Cmathrm%7B~mol%7D%7D%20%5Cchi_%7B%5Ctext%20%7Bpropane%20%7D%7D%3D0.0132%24)
<h3>What is Dalton’s Law?</h3>
Dalton's law of partial pressures is a gas law that states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture. The mole fraction of a given gas in a gas mixture is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gas mixture. This mole fraction can also be used to calculate the total number of moles of constituent gas if the total number of moles of the mixture is known. In addition, the mole fraction can also be used to calculate the volume of a certain gas in a mixtur.
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