Chemistry Assignment: Exploring Gas Laws, Stoichiometry and Kinetics

This chemistry assignment provides solutions to various problems related to gas laws, including calculations involving pressure, volume, temperature, and the ideal gas law. The assignment covers topics such as the kinetic molecular theory, deviations from ideal gas behavior, and stoichiometry. It includes calculations of molar mass, mole ratios, and the determination of molecular formulas. The solutions also address concepts like effusion rates and the relationship between gas properties under different conditions. The assignment utilizes formulas and equations to solve problems and provides step-by-step solutions, referencing relevant chemical principles and laws.

CHEMISTRY ASSIGNMENT
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]

Q1: All gases becomes fluids at certain temperatures and pressure
Q2: As the absolute temperature increases, the average kinetic energy of the molecules
increases. An increase in temperature increases the rate of collision of the particles.
Q3: Low temperatures and high pressures. The deviations of actual gases often agree with the
predictions of the ideal gas equation to a limit of 5% at normal pressures and temperatures.
Q4: The entire above are correct
Q5: E. For the choice C, the kinetic energy refers to the speed of molecules in a given sample
which is often fully dependent on the temperatures. The real molecular composition of the
chemical is not of interest with kinetic energy. Should both chemical be at the same temperature
then their molecules are moving at the same speed and thus their kinetic energy being identical.
Q6: P1V1=P2V2; V2=P1V1/P2
=720×5/790=4.557L
Q7: V1/T1=V2/T2
V2=V1T2/T1
=1×293/303=0.967 L
Q8: PV=nRT
n=PV/RT=2×0.2/0.0821×323
=0.015 moles
Q9: PV=nRT

n=PV/RT=3×2.42/0.0821×373
=0.237 moles
Molar Mass=10/0.237=42.19 g/mol
Q10: N2 (g) + 3H2 (g)  2NH3 (g)
Moles ratio of N: H=1:3
Volume of N=1×20/3=6.667 L
Q11: P1V1/T1=P2V2/T2
V2=P1V1T2/P2T1
=0.5*300×273/1×96
=426.56 mL=4.3×102 mL
Q12: 2Al(s) + 3H2SO4  Al2 (SO4)3 + 3H2
Mole ratio of Al: H2=2:3
Moles of H2 produced=3×0.5/2=0.75 moles
Volume of H2 produced
1 mole=22400 mL
0.75 moles=0.75×22400/1=16800 mL
Q13: P1/T1=P2/T2

P2=P1T2/T1=1.11842×723/293
=2.75 atm
Q14: PV=nRT
n=PV/RT=0.470×0.1/0.0821×400
=0.00143 moles
Molar Mass=0.1/0.00143=69.93g
Molecular Formula=n×Empirical formula
n=MF/EF=69.93/14
=4.995≅ 5
Molecular Formula= (CH2)5
=C5H10
Q15: PV=nRT
V=nRT/P=1.33×0.0821×300/1.15
=28.485 mL
Q16: PV=nRT
V=nRT/P; n=m/MM
V=mRT/MMP

V/m=MMP/RT
m/V= (32×0.85)/ (0.0821×373)
=27.2/30.6233=0.888g/L
Q17: Mole ratio of CaC2:C2H2=1:1
Moles of C2H2
PV=nRT; n=PV/RT
=10*20/ (0.0821*293)
=8.3147 moles
Moles of CaC2=8.3147 moles
Mass of CaC2=8.3147×64
=532.11g
Q18: Moles of N=63.64/14=4.5457 moles
Moles of O=36.36/16.01=2.2711 moles
Simplest Form of the moles
N=4.5457/2.2711=2
O=2.2711/2.2711=1
Empirical formula=N2O

Molecular Mass=
196 mL=0.385 g
22400 mL=22400×0.385/196=44 g
Molecular Formula=n× Empirical formula
n=MM/EM=44/44
Molecular Formula= (N2O) 1= N2O
Q19: (P+ a
V 2 )
Q20: Rate of effusion NOF/HBR= (M.W.HBR/M.W. NOF) 1/2= (80.912/49.005)1/2
=1.28:1
Q21: A. The volume of the gas particles is much smaller than the distance between the gas
particles. At relatively high pressure, the values of PV/RT tend to be lower than the ideal and at
extremely high pressure PV/RT values are often greater than ideal.
Q22: Relationship between Hg and atm
1 atm=30 Hg
19 Hg=19/30=0.633 atm
Q25: P1V1T2=P2V2T1
T2=P2V2T1/P1V1= (2.50×435×355)/ (1.88×28.5)

Final temperature=720.53K
Q26: A
Q27: PV=nRT
Moles of the halogen, n=PV/RT
=1.41*0.109/ (0.0821×398) =0.004706 moles
Molar Mass=0.334/0.004706=71.0 g/mol
This corresponds to the molar mass of Cl2
Q28: PV=nRT
PV= (m/MM)/RT
MM=mRT/PV=0.465×0.0821×298/ (1.22×0.245)
=38.0 g/mol
Q29: No. of atoms
100 g contains 30.45g N and 69.55 g O
Moles of N=30.45/14.01=2.173 moles
Moles of O=69.55/16.014.347 moles
Simplest form of the Moles
N=2.173/2.173=1

O=4.347/2.173=2
Empirical formula=NO2
PV=nRT
n=PV/RT= (1.02×0.389)/ (0.0821×273)
=0.0177 moles
Molar Mass=1.63/0.0177=92.1 g/mol
Molecular Formula =n× Empirical formula
n=MF/EF=92.1/46.01
=2
Molecular Formula= (NO2)2= N2O4
Q30: 50.0 g KClO3=50.0 gKClO3/122.55g KClO3
=0.40799 moles
2 moles of KClO3 will generate 3 moles of O2
Hence, 0.40799 moles of KClO3
0.612×0.40799×2/3 moles of O2
At STP, 1 mole of gas=22.4 L
Set up ratio

0.612 moles of O2/x litres=1 mole/22.4L
x=13.7 L of O2
Q31: D. All molecules have kinetic energy. The heaviest will hence move the most slowly

References
Alezi, D., Belmabkhout, Y., Suyetin, M., Bhatt, P. M., Weseliński, Ł. J., Solovyeva, V., ... &
Eddaoudi, M. (2015). MOF crystal chemistry paving the way to gas storage needs:
aluminum-based soc-MOF for CH4, O2, and CO2 storage. Journal of the American
Chemical Society, 137(41), 13308-13318
Atkins, P., De Paula, J., & Keeler, J. (2018). Atkins' physical chemistry. Oxford university press
Chang, R., & Overby, J. (2000). General chemistry: the essential concepts. Mc Graw Hill
Haynes, W. M. (2014). CRC handbook of chemistry and physics. CRC press
Silberberg, M. S. (2007). Principles of general chemistry (p. 29). New York: McGraw-Hill
Higher Education