Question 1 (a),(b) and (c) of Gar Constant Post-Lab Questions 1. An evaluation o

Question

Question 1 (a),(b) and (c)

of Gar Constant Post-Lab Questions 1. An evaluation or R was performed udng the procedures collected was 32.7 was and the lem was 2 The volume of HaE) mL, The perature 0.0KITL value of R was Laumimol a) How many grams or M metal were used in this deerminati (b) f the correction for the vapor presaare of water had rot been performed, whal would be the calculated value R? (c) If the syringe volume had been incorrectly read resalning in a d H. volume of 30.6 ml., what would be the percent emor n the calculated value of R? 2. (a calculate the volume of one mole of hydrogen gas at 273 k and 160 torr, dUse a value of R 0.0821 L atm mol K) (b) Would you expect the volume of one mole of oxygen gas under the same conditions to be the same, or different? Explain. 3. of an ideal gas originally at 380 torr and 298 K are compressed at constant temperature to a final pressure of 680 torr. Use the ideal gas law to calculate the volumes of gas at the initial and final pressures. Initial volume (a 380 torr Final volume (@680 torr):

Explanation / Answer

aa)

mass of Mg used

relate via:

Mg(s) + 2H+(aq) --> H2(g) + Mg+2

mol of H2 -->

PV = nRT

Pgas = Ptotal - Pvapor = 735-19.3 mm Hg = 715.7 mm Hg of Hydrogen gas

n = PV/(RT)

n = (715.7)(32.7*10^-3)/(62.3*(21.5+273))

n = 0.001275 mol

ratio is 1 mol of MG per 1 mol of H2

0.001275 mol of MG then

mass = mol*MW = 0.001275*24.3 = 0.0309825 g of Mg

b)

if we do not correct:

PV = nRT

R = PV/(nT)

R = (735)(32.7*10^-3)/(0.001275*(21.5+273))

R = 64.0 --> 64.0/760

R = 0.08421052 Latm/moK

c)

V = 30.6 mL recalculate R:

R = PV/(nT)

R = (715.7 )(30.6*10^-3)/(0.082*21.5+273))

R = 0.07970 Latm/molK

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