11) Calculate the volume, in liters, occupied by each of the following: a. 2.00 moles of H₂ at 300. K and 1.25 atm. 39.4 L b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C 14.9 L

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Would you help me with the number 1 question? Also, can you show me the step-by-step and formula?
.
5
Initial Pressure (P₁) = 12 atm
Initial volume (V₁) = 23L
Final Pressure (Pa)
Final volume V
Initial temperature
Use the combined
gas
- 300k
law to solve
the following problems: temperature (T₁) = 200K
8) If I initially have a gas at a pressure if 12 atm, a volume of 23 liters, and a temperature of 200.0 K, and
then I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of the
gas?
V₂ = Pivita
P₁V₁
V₂-12am x23LX300.0k
Pa Va
Ta
14 atm x 200.0k/
тра
29.57142857
BOL
9) A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the
temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?
Final volume (Va)
·P,V,
Pa Va
30. L
31 L
TI
v₂= P₁Vita= 2,3atmx 17LX350k
Ideal Gas Law PXT
1.5 atm x 2,99 K
v₂ =
V=312]
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pnrt R=0.0821 1 atm mo [¹
molen = 1.35 mole Temperature (1): 320k P = nr I
volume v= 250L Pressure (1):
.14 atm
b. 4.75 L of gas containing 0.86 moles at 300, K.
volume(v): 4.75L mole:0.86 mole
Temperature: 300k Pressure:
4.5 atm
initial volume (V₁) = 172
initial pressure (P₁) = 2.3 atm
initial temperature (T) 299K
39.4 L
14.9 L
P= nrt
V
P= 4.459 atm
(4.5 atm)
11) Calculate the volume, in liters, occupied by each of the following:
a. 2.00 moles of H₂ at 300. K and 1.25 atm.
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
.0646 moles
Final Pressure (P₂) = 1.5atm
Final temperature (1₂)= 350 k
30.51282051
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C
.030 moles
= 1.35 mol x 0.082 1 atm mol kx320k
250L
P=0.14 atm
= 0.86m 1X0.0821 Latm mol¹k X 300k
4.75 L
12) Determine the number of moles contained in each of the following gas systems:
a. 1.25 L of O₂ at 1.06 atm and 250. K
op
Transcribed Image Text:. 5 Initial Pressure (P₁) = 12 atm Initial volume (V₁) = 23L Final Pressure (Pa) Final volume V Initial temperature Use the combined gas - 300k law to solve the following problems: temperature (T₁) = 200K 8) If I initially have a gas at a pressure if 12 atm, a volume of 23 liters, and a temperature of 200.0 K, and then I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of the gas? V₂ = Pivita P₁V₁ V₂-12am x23LX300.0k Pa Va Ta 14 atm x 200.0k/ тра 29.57142857 BOL 9) A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas? Final volume (Va) ·P,V, Pa Va 30. L 31 L TI v₂= P₁Vita= 2,3atmx 17LX350k Ideal Gas Law PXT 1.5 atm x 2,99 K v₂ = V=312] 10) Calculate the pressure, in atmospheres, exerted by each of the following: a. 250 L of gas containing 1.35 moles at 320 K. Pnrt R=0.0821 1 atm mo [¹ molen = 1.35 mole Temperature (1): 320k P = nr I volume v= 250L Pressure (1): .14 atm b. 4.75 L of gas containing 0.86 moles at 300, K. volume(v): 4.75L mole:0.86 mole Temperature: 300k Pressure: 4.5 atm initial volume (V₁) = 172 initial pressure (P₁) = 2.3 atm initial temperature (T) 299K 39.4 L 14.9 L P= nrt V P= 4.459 atm (4.5 atm) 11) Calculate the volume, in liters, occupied by each of the following: a. 2.00 moles of H₂ at 300. K and 1.25 atm. b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C .0646 moles Final Pressure (P₂) = 1.5atm Final temperature (1₂)= 350 k 30.51282051 b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C .030 moles = 1.35 mol x 0.082 1 atm mol kx320k 250L P=0.14 atm = 0.86m 1X0.0821 Latm mol¹k X 300k 4.75 L 12) Determine the number of moles contained in each of the following gas systems: a. 1.25 L of O₂ at 1.06 atm and 250. K op
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