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o-na [289]
3 years ago
11

In an equimolar mixture of o2 and n2, are the mass fractions equal?

Chemistry
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

No

Explanation:

The mass fraction is defined as:

w_{i}=\frac{m_{i} }{m_{t} }

where:

  • wi: mass fraction of the substance i
  • mi: mass of the substance i
  • mt: total mass of the system

<u><em>The mass fraction of two substances (A and B), will be the same, ONLY if the mass of the substance A (mA) is the same as the mass of the substance B (mB).</em></u>

An equimolar mixutre of O2 and N2 has the same amount of moles of oxygen and nitrogen, just to give an example let's say that the system has 1 mole of O2 and 1 mole of N2. Then using the molecuar weigth of each of them we can calculate the mass:

mA= 1 mole of O2 * 16 g/1mol = 16 g

mB=1 mole of N2 *28 g/1mol=28 g

As mA≠mB then the mass fractions are not equal, so the answear is NO.

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Answer:

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Which<br> countries were the world leaders in space exploration throughout the 1960's?
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the us and the soviet union

Explanation:

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Determine the mass of a ball with a wavelength of 3.45 × 10-34 m and a velocity of 6.55 m/s
Yuki888 [10]
<span>293 grams The formula for the wavelength of a massive particle is λ = h/p where λ = wavelength h = Plank constant (6.626070040Ă—10^â’34 J*s) p = momentum (mass times velocity) So let's solve for momentum and from there get the mass λ = h/p λp = h p = h/λ Substitute known values and solve p = 6.626070040Ă—10^â’34 J*s/3.45Ă—10^-34 m p = 1.92 J*s/m Since momentum is the product of mass and velocity, we have p = M * V p/V = M So substitute again, and solve. p/V = M 1.92 J*s/m / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 0.293 kg = M So the mass is 293 grams</span>
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A student sets up an experiment according to the table below. What question was the student most likely investigating?
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Explanation:

7 0
3 years ago
A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of
ipn [44]

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L

7 0
3 years ago
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