The answer is 1. A sample of 0.50 moles of gas is placed in a container of volume of 2.5 L. What is the pressure of the gas in torr if the gas is at 25oC? P = 4.89 atm = 3719 torr
2. A sample of gas is placed in a container at 25oC and 2 atm of pressure. If the temperature is raised to 50oC, what is the new pressure? P = 2.17 atm
3. At 1 atm of pressure water boils at 100oC, if the sample was placed under 2 atm of pressure, what would be the temperature? (This would be like a pressure cooker).
T = 746 K = 473oC = 883oF
4. At what temperature would water boil if the pressure is 600 torr? (Use information from problem 3: this shows why food doesn't cook well at higher elevations)
T = 294 K = 21.5oC = 70.7oF
5. Calculate the volume of 40.6 g of F2 at STP. V = 23.9 L
6. A sample of 2.0 moles of hydrogen gas is placed in a container with a volume of 10.4 L. What is the pressure of the gas in torr if the gas is at 25oC? P = 4.70 atm = 3576
7. The tire pressure is 32 psi. What is the pressure in torr if 1 atm = 14.7 psi?
P = 1654 torr
8. A gas is placed in a balloon with a volume of 3.0 L at 28oC and 900 torr. What would be the new volume for the gas if placed under STP? V = 3.2 L
9. How many moles of gas would occupy a volume of 14 L at a pressure of 700 torr and a temperature of 30oC? n = 0.52 mol
10. Calculate the volume of 24.0 g of HCl at STP. V = 14.8 L
11. What is the volume of one mole of acetylene gas at STP? V =22.414 L
12. What is the volume of 0.75 mol of gas at 72oC and 2 atm? V = 10.6 L
13. After eating beans, a student collects a sample of gas at 0.97 atm and 26oC which occupies a volume of 3.5 L, calculate its volume at STP. V = 3.1 L
14. Ammonia (NH3) is placed in 1.5 L flask at 25oC. If the pressure of the gas is 0.899 atm, what is the density? d = 0.626 g/L
15. A mixture of Ar and CO gases is collected over water at 28oC and an atmospheric pressure of 1.05 atm. If the partial pressure of Ar is 600 torr, what is the partial pressure of CO? (vapor pressure of water at 28oC is 28.3 mmHg) PCO = 0.223 atm
16. Determine the partial pressures of each of the gases in the following mixture: 17.04 g NH3, 40.36 g Ne and 19.00 g F2. The gases are at 1.5 atm of pressure.
PNH3 = 0.428 atm; PNe = 0.857 atm; PF2 = 0.2124 atm
17. Potassium chlorate decomposes under heat as follows:
2 KClO3 (s) -------> 2 KCl (s) + 3 O2 (g)
The oxygen gas is collected over water at 25oC. The volume of gas is 560 mL measured at 1 atm. Calculate the number of grams of KClO3 used in the reaction. (vapor pressure of water = 0.0313 atm) nO2 = 0.022 mol; 1.81 g KClO3
Answer:
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Explanation:
CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)
CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0
ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0
S°(J/K·mol): 186.1 188.8 213.7 130.7
ΔG<0 to be spontaneous
ΔG = ΔH- TΔS <0
ΔH = ∑nΔH(products) - ∑nΔH(reactant)
ΔH = (-393.5) - (–74.87 + 2*–241.8)
ΔH = 164.97 kJ = 164970 J
ΔS = ∑nΔS(products) - ∑nΔS(reactant)
ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)
ΔS = 172.8 J
0 > 164970 J - T* 172.8 J
-164970 J > - T* 172.8 J
954.7< T
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Answer:
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of NaOH = x
Molar mass of NaOH = 40 g/mol
Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)
Molarity of the solution of NaOH = 0.250 M
Solving for x:
x = 4.5 g
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Answer:
7.00
Explanation:
When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.
First, let's find the number of moles of each one of them, multiplying the concentration by the volume:
nH+ = 0.15 M * 25 mL = 3.75 mmol
nNH3 = 0.52 M * 25 mL = 13 mmol
So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.
