Gas Laws Boyle\'s Law, and Determining the ideal Gas Constant R Show all calcula
ID: 901045 • Letter: G
Question
Gas Laws Boyle's Law, and Determining the ideal Gas Constant R Show all calculations. Report your answer in scientific notation with correct significant figures and units. R = 0.08206 L atm mol K If 50 ml. of oxygen gas is compressed from 20 atm to 40 atm of pressure, what is the new volume at constant temperature? What volume will be occupied by one-tenth of a mole of an ideal gas at 25degreeC and 760 tort? A student obtained the following data following for collecting the sulfur dioxide gas over water from the decomposition of sulfurous acid (density = 8.62 lbs/gal). Temperature of water, degreeC 45 Barometric pressure, inches of Hg 30.45 Volume of the gas collected. mL 532.14 Write and balance the molecular equation for the decomposition reaction. How many milliliters sulfurous acid were decomposed? A student obtained the follow ing data follow ing the procedure of this experiment. Weight of flask, foil cap, unknown, g 91.596 Weight of flask, foil cap, g 91.027 Temperature of boiling water, C 100 Barometric pressure, mm Hg 784 Volume of the flask, mL 212 Calculate the density and molar mass of the vaporized unknown gas.Explanation / Answer
1. When the temperature is kept constant, we can derive the equation:
P(1) x V(1) = P(2) x V(2)
Where:
P(1) is the pressure of gas before compression
V(1) is the internal volume of the cylinder
P(2) is the pressure of gas after compression
V(2) is the volume of gas at pressure P (2)
So, the new volume = (50*20)/40= 25 ml
2. 1 mole of an ideal gas at STP is 22.4 litre. So, the volume of 1/10th of an ideal gas at 250C and 750 torr is 0.1*22.4= 2.24 litre.
3. Balanced molecular equation for the decomposition reaction: H2SO3 <==> H2O + SO2
So, using ideal gas equation:
moles of SO2 formed = PV/RT = (30.45*0.0334)atm*0.532L / (0.0821 L atm K-1 mol-1*318 K) = 0.0207
Now, according to balanced chemical equation: moles of H2SO3 decomposed = moles of SO2 formed.
Thus, moles of H2SO3 decomposed= 0.0207
mass of H2SO3 decomposed= 0.0207*82.07 = 1.6988 g
Volume of H2SO3 decomposed = mass/density= 1.6988g / (8.62*0.119826) g/ml = 1.645 ml
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