Question

Please answer all the blanks (volume if H2 and everything in analysis). TIA!

Data 5 1 oong 0.00 10.5ml 2 o.olag 0.00 11.0 Trial 3 o.org 0.00 12.00 o Daag o.albg 0.00 10.0 ml 11.5ml Mass of Mg (g) Initial volume of Syringe (mL) Final volume of Syringe (mL) Volume of H (mL) Barometric pressure (torr) Ambient temperature (°C) Vapor pressure of H2O (torr) 779.314har 23. Oi 21.0 forr TA.314tar 23.0c 179.3 14ton 23.0¢ 779.314 ton 23.0c 779.31472 23.0c 21.0 torr 2.Oferr 210 21.0torr Analysis Your goal is to determine the value of R from your data. Perform each of the following operations. Attach a page showing your work. You do not have to show work for every trial, but show your work for one sample trial for each step. 1. Determine the number of moles of magnesium consumed in the reaction, which will be equal to the number of moles of H2 produced (see Equation (7-2)). Moles of Mg (g) 2. Use Equation (7-3) to determine the pressure of H, in the sample. Express your answer in atmospheres. Pressure of H2 (atm) 3. Convert the ambient temperature to Kelvin. Ambient temperature (K) 4. Use Equation (7-1) to calculate R. Remember to convert mL to L to get the units below. R (L atm/(mol K)) 5. Calculate the average and standard deviation of your values of R. Average R (L atm/(mol K)) Standard deviation in R (L atm/(mol K)).

Answer 1

- 7.82x10 Trial - reass of mag ug = 0.0199 atomic mass of reg= 24.3 . No. of moles of My oota 0:019 24.3 - 2tlar) + reges) regely (as) + 142 19 Imole regis) produced Imole litg . d. No. of moles of the produced = 7.82x10 mole Barometric pressure = 779.31 tors = 2 - 779.31 - 1.025 atm vapour pressurot H2O = 210 torr = 21.0 = 0.028 atm As Patma Pitz & PHO Pite = Patm - Piago = 1.025 0.028 = 0.997 atm 760 760 Ambient temperature - 23+273 = 296k

Initial volumest final volumet syninge=0.000 syringe = 10.5 ml - volume of Hoa V-via 10.5-0= 10.5 mL. -0.01052 from er 7.1 PV = nRT PV 0.997 X 0.0105 = 0.0452 Latul T = 7.82x104 x 296 molk Trial - 2 Mass of reg = 0.019 g. No of moles of meg a 0.019 - 7.82x10 mole 24.2 initial volume = 0.com final volume 11.00 ml - volumest its a 11.00ml = 0.0110L Barometric pressure. 779.314 torr = 779.314 = 1.025-4 atm 760 Vapour pressur of ho= 21.0 torra 0.028 atm .: PH2 = Patm- Plogo = 1.0254-0.028 = 0.9974 atin now Poo e Pov . 0.9974 x 0,01 782x104 x 296 = 0.0474 Lahm,malkt Trial - 3 mass of reg = 0.017g. No. of mole of My = 0.017 = 6,996410" mole volume of to a 12.0-0.0 = 12.0 mL = 0.012 L Baronatric Pressure - 1.0254 atin vapour pressur of logo o 0.028 atin MT 24.3

0.997u R= Pressure at ito a 1.0256 - 0.028 = 0.9974 aton 0.9974 x 0.012 6.996410 4 x 296 - 0.0578 Lahm molk! Torial- 34 Mass of lege 0.016 f. No. of moles of suga 0.0166.584810 "mole = moles of the formed volume of to = 100.0o = 10.0mL = 0.010L Pity = 0.9974 atm; T = 296k R-PV = 0.9974 x 0.010 - 0.0512 L.atm. motke 6.5848104 x 296 7 24.3 NE Trial -5 24.3 Mass of reg= 0.022g No. of moles of ug 0.022 -9.083x10 mole= moles of 172 formd rolume of it2 = 11.5-0. 0 1 15mL = 0.01152 PH₂ = 0.9974 ahm, T = 2966 0.9974 X 0.0115 -0.0428 L.atm. motkh 9,053410 4x296 - Average R = 0.0452 +0.0474 +0.0578+ 0.0512 +0.0428 = 0.0489 L.atm.mollkt PV R? NT 2 Trial 3 4 5 mass of reg (9) 0.019 0.019 0.017 0.016 0.022 vol of H₂ L) 0.0105 0.0110 0.012 0.010 0.0115 moles of reg 7, 821104 7.828104 6,996x104 6.5848104 9.0530104 Pressure of Ha fatur) 0.997 0.9974 0.9974 0.9974 0.9974 296 296 296 temperature 296 296 IK) 0.0512 0.0428 0.0474 0.0578 (2.atm/molek) 0.0452