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saw5 [17]
3 years ago
14

A 7.70 L 7.70 L container holds a mixture of two gases at 47 ° C. 47 °C. The partial pressures of gas A and gas B, respectively,

are 0.344 atm 0.344 atm and 0.893 atm. 0.893 atm. If 0.190 mol 0.190 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

1.884 atm

Explanation:

We have at first, a mix of two gases A and B, with a pressure and temperature. The total pressure of this mix is the sum of both pressures fo these gases so:

P = 0.344 + 0.893 = 1.237 atm

Now after this, a third gas is added but there is no change in temperature or volume, which means that this gas may have a behavior of ideal gas, therefore, we can use the expression for ideal gas which is:

P = nRT/V

R is the gas constant, in this case is 0.082 L atm/ K mol

Replacing all the values that we have of volume and temperature, we can calculate the pressure that this gas exert:

P = 0.19 * 0.082 * (47+273.15) / 7.7

P = 0.647 atm

Now the total pressure, we add this value and we have:

P' = 1.237 + 0.647

P' = 1.884 atm

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2 years ago
According to the following balanced equation, 2 formula units of Iron (III) Oxide (Fe2O3) can be formed by reacting 4 atoms of i
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You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
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6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol <span>O₂
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amount of Fe₂O₃ produced:
<span>(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe</span>₂O₃<span> 
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6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
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</span>1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
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<span>(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
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I hope this helps.
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