Click to see additional instructions A balloon at Earth's surface is filled with 100% pure oxygen. The balloon and the environment have the same pressure, Penv = Pb = 1013 hPa and the environmental air temperature (Tenv) is 25°C. Calculate the temperature the Oxygen in the balloon needs to have in order to NOT be buoyant. Hint: Think about the terms in the ideal gas law. Balloon's Temperature to be Neutrally Buoyant = 77.98 °C
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Problem Set #3
Q11
I got 77.98 C,
If I am wrong for the temperature, please how to get the right answer. Ty
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- The pressure exerted on the gas entrapped in the syringe is proportional to the total mass on the platform. Make two plots. (i) Vol/mL vs. Total mass/g (ii) Vol/mL vs. 1/(total mass/g) If warranted, obtain a trendline and trendline equation. Comments in the results. Is Boyles Law obeyed? Explain.At 27 °C, 500 cc of H2, measured under a pressure 400 torr, and 1000 cc of N2, measured under a pressure of 600 mm Hg, are introduced into an evacuated 2-L flask. Calculate the resulting pressure. (Given -> Unknown -> Solution -> Answer)An open-end mercury manometer is to be used to measure the pressure in an apparatus containing a vapor that reacts with mercury. A 10-cm layer of silicon oil (SG = 0:92) is placed on top of the mercury in the arm attached to the apparatus. Atmospheric pressure is 765 mm Hg.(a) If the level of mercury in the open end is 365 mm below the mercury level in the other arm, what is the pressure (mm Hg) in the apparatus?(b) When the instrumentation specialist was deciding on a liquid to put in the manometer, she listed several properties the fluid should have and eventually selected silicon oil. What might the listed properties have been?
- Calculate the molar volume of water at 50 ºC and 1 atm, and at 50 ºC and100 atm.Data: ρ50 ºC, 1 atm = 0.98804 g/cm3; κ = 4.4•10-10 Pa-1.we use Pv=nRT. Everything here should be related to H2 (g) --> Chemical equation1Mg +2 HCl à MgCl2 +H2(g) From video : mass of Mg =.0407 g H2 gas volume = 38.2ml Temperature of room=24.0 +273=297k Total Pressure of the room = 30.01 in mmHg Calculations: find total pressure from barometer and convert it to mmHg, 1 in =2.54 cm and 25.4 mm 30.01 in mmHg x 25.4 mm Hg =762.3 mmHg total pressure of room 1in mmHg First step- convert g of Mg to mol of H2 0.0407g Mg x 1mol/24 gx 1molH2 = 0.00169mol 1mol Mg Second step: Total pressure of gas = Pressure of H2 + P of water 762.3 = p H2 + 22.4 FROM reference (canvas)=Pressure of water at 24 C= 22.4 mmHg n= 0.00169mol H2 Pressure of H2= 762.3-22.4= 739.854 x 1atm/760= 0.973 atm H2 Volume of H2= 38.2 ml /1000 =0.038L H2 Temp= 297K anwer this quesions with the information provide: Using PV=nRT, Find R=? (No R averge needed) à 0.973X0.0382=0.00169XRx 297 0.0372=0.50R Then, Percent error…An experiment is conducted in which a quantity of H2(g) is chemically generated in a closed headspace of volume V1 at absolute temperature T1, thereby raising the total pressure from Pa (the initial air pressure) to Pf (the total final pressure of H2(g) + air). Compose an algebraic expression for the volume (V2) that this sample of H2(g) would occupy if it were isolated (pure -- i.e. no air) at absolute temperature T2 and partial pressure P2. Note that your expression may contain no other variables than those given above, namely: V1, T1, T2, Pa, Pf, and P2. A) V2 = Now, suppose that the above experiment generated n mol of H2(g), that T2 = 0 °C and that P2 = 1 atm. Give the simple algebraic expression, in terms of V2 and n, which equates to the molar volume of H2(g) at STP. B) VM =
- An experiment is conducted in which a quantity of H2(g) is chemically generated in a closed headspace of volume V1 at absolute temperature T1, thereby raising the total pressure from Pa (the initial air pressure) to Pf (the total final pressure of H2(g) + air).Compose an algebraic expression for the volume (V2) that this sample of H2(g) would occupy if it were isolated (pure -- i.e. no air) at absolute temperature T2 and partial pressure P2.Note that your expression may contain no other variables than those given above, namely: V1, T1, T2, Pa, Pf, and P2. V2 = Now, suppose that the above experiment generated n mol of H2(g), that T2 = 0 °C and that P2 = 1 atm. Give the simple algebraic expression, in terms of V2 and n, which equates to the molar volume of H2(g) at STP.VM = 8.829×10-4 mol of B2H6(g) (diborane gas) is generated into a 176.0 mL headspace, at 25.0 °C, thus increasing the pressure by 0.1209 atm. If we assume that the diborane gas obeys Boyle's Law and Charle's Law, what…An experiment is conducted in which a quantity of H2(g) is chemically generated in a closed headspace of volume V1 at absolute temperature T1, thereby raising the total pressure from Pa (the initial air pressure) to Pf (the total final pressure of H2(g) + air).Compose an algebraic expression for the volume (V2) that this sample of H2(g) would occupy if it were isolated (pure -- i.e. no air) at absolute temperature T2 and partial pressure P2.Note that your expression may contain no other variables than those given above, namely: V1, T1, T2, Pa, Pf, and P2.Dry ice is sometimes used for special effects to form a "fog" as it sublimes from the solid to the gaseous state. If 99.0 grams of dry ice is used to generate this "fog", how many torr of pressure will be exerted by the gaseous carbon dioxide once all of the dry ice sublimes? Assume the volume of the auditorium where the fog is generated is 1.24 x 10^{5}5 m^{3 }3and the temperature is 22ºC. (Enter your answer using either standard or scientific notation. For scientific notation, 6.02 x 10^{23}23 is written as 6.02E23.)
- At 27 °C, 500 cc of H2, measured under a pressure 400 torr, and 1000 cc of N2, measured under a pressure of 600 mm Hg, are introduced into an evacuated 2-L flask. Calculate the resulting pressure. Please help asapHere in this PV=nRT problem, I solve for the temperature in C. 800 mm Hg / 760 mm Hg = 1.052631579atm. Chlorine gas is cl2, 100g/70.9g=1.410437236. After algebra, I get 227.211655. Is my answer off because I needed to round somewhere?1. The compressibility factor Z = PV/nRT for O2 at -27.1°C and 537.1 atm is 1.50; at 147.4°C and 95.5 atm, it is 1.02. A certain mass of oxygen occupied a volume of 3.28 L at 147.4°C and 95.5 atm. Calculate the volume occupied by the same quantity of oxygen at -27.1°C and 537.1 atm. 2. The composition of a mixture of gases in percentage by volume is 20.5% nitrogen, 15.7% carbon monoxide, 37.8% water, and the rest sulfur dioxide. Calculate the density of the gaseous mixture at 39.8°C and 1.46 atm in grams/liter. Report your answer to the hundredths position.