1) A sample of 513 L of Kr gas at 225 K in a container pressurized at 1.562 atm is compressed to 227.4 L with a pressure of 2.08 atm. The temperature needed to maintain these conditions would be ______________ K.
2) For the reaction shown below, write the total ionic equation. Then write the net ionic equation.
CrBr6 (aq) + 3 CaS (aq) -----------------------------> CrS3 (s) + 3 CaBr2 (aq)
3) If 1.776 kg of CrBr6 was dissolved in 4011 mL of water the molar concentration of the CrBr6 solution would be ________
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1) A sample of 513 L of Kr gas at 225 K in a container pressurized...
A 10.33 mol sample of krypton gas is maintained in a 0.7797 L container at 301.0 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L’atm/mol and b = 3.978x10-2 L/mol. atm
A 9.450 mol sample of krypton gas is maintained in a 0.8100 L container at 300.1 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. atm
A 10.13 mol sample of krypton gas is maintained in a 0.7517 L container at 297.2 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol.
A 10.83 mol sample of argon gas is maintained in a 0.7598 L container at 295.6 K. What is the pressure in atm calculated using the van der Waals' equation for Ar gas under these conditions? For Ar, a = 1.345 Llatm/mol2 and b = 3.219x10-2 L/mol. atm USE LIIC CICICULUS LU al USS IITIPUI lalil va The rate of effusion of H2 gas through a porous barrier is observed to be 1.17 x 10-4 mol/h. Under the same conditions,...
A container containing 6.19 L of a gas is collected at 244 K and then allowed to expand to 23.4 L. What must the new temperature be in order to maintain the same pressure? Express your answer in units of K. You do not need to enter the unit. When 16.8-grams of an unknown compound is placed in 0.176-kg of benzene, the freezing point is changed by 1.93oC. The value of Kf= 4.90oC/m for benzene. Determine the molar mass of the...
1. A 10.80 mol sample of oxygen gas is maintained in a 0.8395 L container at 304.6 K. What is the pressure in atm calculated using the van der Waals' equation for O2 gas under these conditions? For O2, a = 1.360 L2atm/mol2 and b = 3.183×10-2L/mol. ______atm 2. According to the ideal gas law, a 1.093 mol sample of nitrogen gas in a 1.390 L container at 266.8 K should exert a pressure of 17.22 atm. What is the...
A 1.55-mol sample of nitrogen gas is maintained in a 0.730-L container at 292 K. Calculate the pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals constants for N2 are a = 1.39 L2atm/mol2 and b = 3.91×10-2 L/mol). Pideal gas equation = ______ atm Pvan der Waals =_____ atm
w4.
1a) For the following reaction, 0.275 moles of
hydrochloric acid are mixed with
0.372 moles of oxygen gas.
hydrochloric acid (aq) + oxygen (g) --> water (l) + chlorine
(g)
What is the formula for the limiting
reagent?
What is the maximum amount of water that can be
produced? (answer in moles)
1b) An iron nail rusts when exposed to oxygen.
For the following reaction, 0.343 moles of
iron are mixed with 0.127 moles
ofoxygen gas.
iron (s) +...
Question 1 (a) Use the ideal gas equation to calculate the pressure (in atm) of 2.40 mol of krypton (Kr) at 455 K in a 4.50 L vessel. (b) In a 16.3 L vessel, the pressure of 2.40 mol of Kr at 455 K is 5.50 atm when calculated using the ideal gas equation and 5.40 atm when calculated using the van der waals equation of state (Note: a=5.121 and b = 0.0106). Why is the percent difference in the...
Hint: % difference = 100×(P ideal - Pvan der Waals) / P idealAccording to the ideal gas law, a 9.843 mol sample of argon gas in a 0.8425 L container at 502.0 K should exert a pressure of 481.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Ar gas, a =1.345L2 atm/mol2 and b = 3.219×10-2 L/mol.