Mole fraction of Oxygen=0.381
Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)
also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)
consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)
= 0.381
The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.
Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.
Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.
The question is incomplete. Please read below to find the missing content.
Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.
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Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of
<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>
This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas
<span><span>7014.04</span>=4.98≈5</span>
Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is
<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>
Answer:
look at the graph
Explanation:
We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.
Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.
So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)
