Question

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Please Repost can't see picture Clearly for all problems show a d label all numbers with 1. (10 points) A 85.0 mL sample of carbon dioxide gas is originally at STP. What pressure will this gas exert, in mm Hg, if the volume changes to 0.408 L while the temperature stays the same? 2. (8 points) What volume, in ml., will 0.0145 g of chlorine gas occupy at a temperature of 131°C and a pressure of 1.95 atm? PV=nRt 1.950=0.0145gx v=0.4 3. (8 points) A 28.0 L sumple of a gas has a mass of 14.6 g at 23.0'C and 1.05 atm. What is the density of the gas at STP?

Answer 1

Answere Boyle's law describes that the pressure exerted by a given mass of ideal gas is inversely proportional to the volume it occupies (Temperature remains constant) pat For comparing the same substance under two different sets of conditions, the law can be usefully expressed as PV, = P2 2 | Initial rolume of coa = 85m Pressure at STP = 760 mm Hg STP means the condition at Stand and Temperature and Pressure. The stand and value of Pressure is taken as I atm ore 760 mm Hg Da Latin

If we put the values in the above equation: Pui= P2V2 => 760 x 85 = P, x 408 (0.4081= 408m) >> P2 = 760x85 408 s P2 = 158.33 mm Hg The final pressure evereted is 158.33 mmHg 8. The gas sample has mass: 14.69 and volume occupied = 282 To calculate Density, we know Density = Hasse Beet, Firest we have to calculate the volume at STP. volume will be changed with change in Pressure and volume but Mass Volume mass won't We need to der to change. determine the volume out STO determine density aut STP.

The Strategy is to use the combined gas Law using the values given in question as ferest set of variables (PI, V , Tid and STP as the second set of variables ( PaVa, To) STP is 0c ore 273.15k and 1 atm pressure and P = 1.05 atm V=281 T, = 23°c ore 296k The combined Gas Law is expressed as Substituting the values into the equation V, P, T, V2 = TIP2 28% 1.05 x 273.15 296 X 1 = 27.13 L. Density at STP : Density at Density Mass / volume = 14.69/27.13€ = 0.538g/1