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. molex0.0821 (2) x 300.0K 39.4 L 1.259tm = 39. 408 39.4L PV=nRT V=nRT P b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C 425 e- IT T-27's 27+0-

Chemistry
10th Edition
ISBN:9781305957404
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...
icon
Related questions
Question
100%

Can you help me with number 11 a.) and number 11 b.)? Were this the correct answer that shows my work includes the significant figures? Also, can you help me with the Ideal gas law formula to plug in?

Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok
Final volume (V₂)
(P) = 12 atm
Pavax Ta
та
Pa
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
V₂ = P gas?
T2
12 atm x23L X300.0K
(200.0k X14atm)
= 29.57142857
30. L
pressure.
Initial volums
Use the combined gas law to solve the following problems:
Ti.Pa
30. L
Piv, Pava Pav₂ x ta
Ta
V2=
V₁. P₁. Ta
Pa Ti
TI
Ta
T₁
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 yolume of the gas?
PIVI Pa Va
Final volume (V₂)
initial volume (V.) = 172
initial Pressure (P₁) = 2.3 atm
T₂
31 L
Final Pressure (P₂) = 1.5 atm
initial temperature (T) = 299k Final temperature (To) = 350k
TaxPixVi
Pa.va.ta
=
=
Pa deal Gas Law
TX.B₂
12=
-XT2 P₁V₁ - P₂V₂
Ta
P.V₁ Ta
=
2.3atm x 77LX 350 k)
11. Sa+m. 299K)
= 30.51282051
31L
Pa Ti
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pv=n·R·T, R=0.0821
Mole n=1.35 mote Temperture (1)=320K P = n.r.r
Volume v= 250L Pressure =
V
.14 atm
P=014 atm
0.1418688
b., 4.75 L of gas containing 0.86 moles at 300. K.
mole: 0.86 mole
Pressure:
14.9 L
volume (v): 4.75L
Temperature: 300k
4.5 atm
V=4.75L
T= 300k
/Leatm
p=4.459atm
n=0.86 mole R=0.0821 (K+mole)
P = 4.5atm
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.
PV=nRT
V=nRT
2.00 mole X0.0821 (2) × 300.0K
1.259tm
39.4 L
P=n·R·T
V
= 39. 408
39.4L
= 1.35 mol x 0.0821 Komole X 320K
250L
= 0.86 molx0.0821 (atm).
4.75X
(2·atm
0.425 mole (NH3) X0.0821 (Komple) X 310.15K_
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C
.030 moles 0.925 atm X 0.80 L
• atm
mole
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
PV=nRT₁ V = nRT n=0.425 mol T=37°C = 37+273.15-k = 310.15 K
atm
P
= 14.94738795
14.92
0.7249tm
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
Pr=nRT = P (1.06 atm x 1.25L)
RT
0.0821 (Leatm/ kmal) X 250.K)
P=1.06 atm R=0.0821 (L.atm/K+mol)
.0646 moles
v1.25L
T=250.K
^=
Komor X 300k
=0.030029646
0.0821L.atm/mol x 300.15K) (030 mole
4.459326316
P= 1.25 atm
V=
T = 300.K
n = 2.001 mot
R=0.082L a+m/ Kime)
= 0.06455542
(.0646 Moles)
PV=nRT
V=0.80L
P=0.925atm
T=276=27+273.15= 30045K
(4.5atm)
= RY
n = PV
Transcribed Image Text:Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok Final volume (V₂) (P) = 12 atm Pavax Ta та Pa 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 V₂ = P gas? T2 12 atm x23L X300.0K (200.0k X14atm) = 29.57142857 30. L pressure. Initial volums Use the combined gas law to solve the following problems: Ti.Pa 30. L Piv, Pava Pav₂ x ta Ta V2= V₁. P₁. Ta Pa Ti TI Ta T₁ 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 yolume of the gas? PIVI Pa Va Final volume (V₂) initial volume (V.) = 172 initial Pressure (P₁) = 2.3 atm T₂ 31 L Final Pressure (P₂) = 1.5 atm initial temperature (T) = 299k Final temperature (To) = 350k TaxPixVi Pa.va.ta = = Pa deal Gas Law TX.B₂ 12= -XT2 P₁V₁ - P₂V₂ Ta P.V₁ Ta = 2.3atm x 77LX 350 k) 11. Sa+m. 299K) = 30.51282051 31L Pa Ti 10) Calculate the pressure, in atmospheres, exerted by each of the following: a. 250 L of gas containing 1.35 moles at 320 K. Pv=n·R·T, R=0.0821 Mole n=1.35 mote Temperture (1)=320K P = n.r.r Volume v= 250L Pressure = V .14 atm P=014 atm 0.1418688 b., 4.75 L of gas containing 0.86 moles at 300. K. mole: 0.86 mole Pressure: 14.9 L volume (v): 4.75L Temperature: 300k 4.5 atm V=4.75L T= 300k /Leatm p=4.459atm n=0.86 mole R=0.0821 (K+mole) P = 4.5atm 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. PV=nRT V=nRT 2.00 mole X0.0821 (2) × 300.0K 1.259tm 39.4 L P=n·R·T V = 39. 408 39.4L = 1.35 mol x 0.0821 Komole X 320K 250L = 0.86 molx0.0821 (atm). 4.75X (2·atm 0.425 mole (NH3) X0.0821 (Komple) X 310.15K_ b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C .030 moles 0.925 atm X 0.80 L • atm mole b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C PV=nRT₁ V = nRT n=0.425 mol T=37°C = 37+273.15-k = 310.15 K atm P = 14.94738795 14.92 0.7249tm 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 Pr=nRT = P (1.06 atm x 1.25L) RT 0.0821 (Leatm/ kmal) X 250.K) P=1.06 atm R=0.0821 (L.atm/K+mol) .0646 moles v1.25L T=250.K ^= Komor X 300k =0.030029646 0.0821L.atm/mol x 300.15K) (030 mole 4.459326316 P= 1.25 atm V= T = 300.K n = 2.001 mot R=0.082L a+m/ Kime) = 0.06455542 (.0646 Moles) PV=nRT V=0.80L P=0.925atm T=276=27+273.15= 30045K (4.5atm) = RY n = PV
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

Blurred answer
Knowledge Booster
Mole Concept
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY