0.447 is the mole fraction of Nitrogen in this mixture.
mole fraction of nitrogen= moles of nitrogen/total moles
mole fraction of nitrogen=0.85/1.90
mole fraction of nitrogen=0.447
The product of the moles of a component and the total moles of the solution yields a mole fraction, which is a unit of concentration measurement. Because it is a ratio, mole fraction is a unitless statement. The sum of the components of the mole fraction of a solution is one. In a mixture of 1 mol benzene, 2 mol carbon tetrachloride, and 7 mol acetone, the mole fraction of the acetone is 0.7. This is computed by dividing the sum of the moles of acetone in the solution by the total number of moles of the solution's constituents:
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Answer:
Alright, the first thing we have to do is to balance the chemical equation
2Na3N -----> 6Na + 1N2
We have 60g of Na3N, we convert them into moles by dividing the mass of the compound by the molar mass.
Molar mass of Na3N = (22.98 x 3) + (14) = 82.94g/mol
<u>60</u> = 0.72341451651 moles of Na3N
82.94
Now because we did the balanced equation, we know the mole to mole ratio of Na3N to N2 would be 2:1, so in order to get the moles of N2 you have to divide the moles of Na3N by 2
0.72341451651 moles/2 = 0.361707258 moles of N2
Now that we have the moles of N2, we just have to determine the mass of it in grams. In order to do that, just multiply the moles by the molar mass of N2 (28g/mol)
0.361707258 x 28 = <u>10.13g of N2</u>
<u>Therefore the decomposition of 60g of Na3N would result in 10.13g of N2 (nitrogen gas)</u>
Answer:
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Explanation:
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Answer:
Δ should be 0.1009
Explanation:
The change in the units volume when temperature change can be expressed as:
∆v = v0Δ
with v0 = the initial volume
with = the volumetric temperature expansion coefficient
with Δ = the change of temperature.
To calculate the final volume vf we'll get:
v = v0 + ∆ = v0(1 + Δ)
The liquid just begins to spill out if v(benzene) = :
v()(1 + Δ) = = v() (1 + Δ)
(v(cavity)-v(benzene))/(((benzene) -(copper)) = Δ
((1.22*10^-3)-(1.1*10^-3))/((1240*10^-6)-(51*10^-6)) = Δ
Δ = 0.1009
Δ should be 0.1009
Answer:
524.82 g Cl2
Explanation:
Balanced Equation:
2Al(s) + 3Cl(g) ⇒ 2AlCl (s)
Al: 26.982 g/mol Cl: 35.453 g/mol
266.28g Al x 1mol Al/ 26.982g Al x 3mol Cl2/ 2mol Al x
35.453g Cl2/ 1mol Cl2 = 524.82 g Cl2
