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I am Lyosha [343]
3 years ago
10

Dinitrogen oxide (N2O) gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet ga

s occupied 123 mL at 21◦C when the atmospheric pressure was 760 Torr. What volume would the same amount of dry dinitrogen oxide have occupied if collected at 760 Torr and 21 ◦C? The vapor pressure of water is 18.65 Torr at 21◦C. Answer in units of mL.
Chemistry
1 answer:
Irina-Kira [14]3 years ago
8 0

Answer:

119.98 mL

Explanation:

Initial volume V1 = 123 ml

Initial temperature = T1 = 21◦C + 273 = 294 K

Initial pressure P1 = 760 Torr - 18.65 Torr = 741.35 Torr

Final volume V2= ???

Final temperature = T2= 21◦C + 273 = 294 K

Final pressure P2 = 760 Torr

From;

P1V1/T1= P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2= 741.35 * 123 * 294/760 * 294

V2 = 119.98 mL

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<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

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solution 1

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<span>C7H8 First, determine the number of relative moles of each element we have and the molar masses of the products. atomic mass of carbon = 12.0107 atomic mass of hydrogen = 1.00794 atomic mass of oxygen = 15.999 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 We have 5.27 mg of CO2, so 5.27 / 44.0087 = 0.119749 milli moles of CO2 And we have 1.23 mg of H2O, so 1.23 / 18.01488 = 0.068277 milli moles of H2O Since there's 1 carbon atom per CO2 molecule, we have 0.119749 milli moles of carbon. Since there's 2 hydrogen atoms per H2O molecules, we have 2 * 0.068277 = 0.136554 milli moles of hydrogen atoms. Now we need to find a simple integer ratio that's close to 0.119749 / 0.136554 = 0.876937 Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio. The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.</span>
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