• ¿Documento equivocado? Cámbialo gratis
  • Escrito por estudiantes que aprobaron
  • Inmediatamente disponible después del pago
  • Leer en línea o como PDF
Vender
¿Dónde estudias?
Tu idioma
Document preview thumbnail
Vista previa 3 fuera de 19 páginas
Examen

CHEM116 Copy of Final Practice Exam Questions and Answers Spring 2026

Document preview thumbnail
Vista previa 3 fuera de 19 páginas

The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS uses for atmospheric pressure are inches of mercury. A barometric pressure of 30.51 inches of mercury corresponds to kPa. A. 16.01 B. 775 C. 103.3 D. 77.50 E. 1.020 A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was atm. 2. A. B. C. D. E. 1.00 16.00 2.00 4.00 8.00 3. A balloon originally had a volume of 4.39 L at 44 °C and a pressure of 729 torr. The balloon must be cooled to °C to reduce its volume to 3.78 L (at constant pressure)

Vista previa del contenido

m >
-
answers

> corrections
m
-




Chem 116 Final Exam Review Sheet PV = NRT

=
soluteae
tmole
Chapter
⑮I업] P1tKe 10
?
. II molality


σ
1. The National Weather Service routinely supplies atmospheric pressure data to help pilots
set their altimeters. The units the NWS uses for atmospheric pressure are inches of
mercury. A barometric pressure of 30.51 inches of mercury corresponds to
kPa. P dgh =
29 92 in
A. 16.01 1 atm =

Ag .




B. 775 latm 101 . 325 kPa . 38639kPa
linHg 3
=
=



g C. 103.3
3 38039
ka
103 . 3 kPa
=
D. 77.50 30 51 inHg ×
,




1 Hg
.




E. 1.020
2. A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at
_
_



constant temperature. After the compression, the gas pressure was atm.
A. 1.00 P V PiVi
,
=




a
(
B. 16.00
C. 2.00 (4 00 atm) ( 8 00λ= ( Pc ) (2 00 l)
.
,
,




D. 4.00
E. 8.00
3. A balloon originally had a volume of 4.39 L at 44 °C and a pressure of 729 torr. The
balloon must be cooled to °C to reduce its volume to 3.78 L (at constant
pressure).


=…
A. 38
B. 0
G C. 273
D. 546 的
E. 72.9
4. The amount of gas that occupies 60.82 L at 31 °C and 367 mmHg is mol.
A. 11.7 PV NRT =

い ℃ )( 60. 87 ) = ( 0 08200) (
. 304 15) n .




σ B. 1.18 4 347 =

mmHg amHy 48 atm
29 1936
= 0 .
n24. 958549 .
=



C. 0.850 n 1 l 7 =
.




D. 894 T 31= 273 15
+ 304 S k
.
=
.
1


E. 0.120
5. At a temperature of °C, 0.444 mol of CO gas occupies 11.8 L at 889 torr.
A. 106
B. 379
C. 32
D. 73
E. 14
6. A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mmHg. The flask is opened and
more gas is added to the flask. The new pressure is 795 mmHg and the temperature is
now 26 °C. There are now mol of gas in the flask.
A. 0.28
B. 2.1
C. 1.6




齒

, D. 3.5
E. 2.9 NOz
7. The mass of nitrogen dioxide contained in a 4.32 L vessel at 48 °C and 141600 Pa is
g. NRt 23 m 014
PV
= 10 58g
= 。
.




4
A. 5.35 × 10 P
B. 9.46 × 10-2
141400 Pax
=
2 gx 10 SPa
4 Oatm
=
に




( 0 08200 ( 321
. 18)
θC. 10.5 ( 4 . 32 ) = . )
40
( ] n

48 273 321 . 15k+ . 15 =
T
n(26 35)
=
.
D. 70.5 . 048
6 =




E. 53.5 23 n =




8. The molecular weight of a gas that has a density of 7.10 g/L at 25.0 °C and 1.00 atm
pressure is g/mol. d = M
=
A. 28.0
7 60) (0 08206) (298 )
15
B. 5.75 × 10-3
.
.

= .




PV nRT =


C. 14.6 709 173
=

.




A
D.
6.85 × 10 -2
d .
RT
=
AN 머쯤
O E. 174 He
9. The volume of hydrogen gas at 38.0 °C and 763 torr that can be produced by the reaction
ー




of 4.33 g of zinc with excess sulfuric acid is L. steps :

mol
① gzn
-
>
-4
A. 2.71 × 10 zn(s)
HzS0y(a9) znSOy(aq) Hz(9)
+ -
> +
atm
② convert K
toor >
imol
-




B. 0.592 molzn +. 33g X55 mol
=
= 0 06623
389
.




. ℃ 5 →



C. 2.84 MRT PV =
③ plug PV nRT in
=




eD. 1.69 004 V
) = 10 1 (
( 0 08200} ( )
, 006233 . , 311 . 15


E. 3.69 × 104 V 1 68 = .




10. Ammonium nitrite undergoes thermal decomposition to produce only gases:
NH4NO2 (s) → N2 (g) + 2H2O (g)
_

What volume (L) of gas is produced by the decomposition of 35.0 g of NH4NO2 (s) at

Xmol 3m
e
525 ° C and 1.5 atm? 35. 0g X
A. 15 x , NHyNo
1 64ms =
.




σ B. 72 (1 . 5) v ( 64)(0 08206) (798 15)
=
.

. .




C. 160 1 3 √ .
107 ll = .




D. 24 √= F 1 6 し .




E. 47
11. In a gas mixture of He, Ne, and Ar with a total pressure of 8.40 atm, the mole fraction of
Ar is if the partial pressures of He and Ne are 1.50 and 2.00 atm,
respectively. 4 9 XAr ( 8 40) = .




A. 0.357
.




× 583 = 0
ar
.




B. 0.417 408
. 1. 50 2 00 - .9 -
.
= 4



∞ C. 0.583
D. 0.238
E. 0.179
12. The root-mean-square speed of CO at 113 °C is m/s.
A. 31.5
B. 317 Vrms Wo =




σ C. 586 R =8 31 서 .
惑弘 O8
=




D. 993 T 380 15 k 586 39
=
=


E. 58.3
.


.




M = . 01
28
glmd
=. 0280 kg/mol

, 13. A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take
s for the same amount of CO2 to effuse under the same conditions.
A. 0.23

⑨
B. 3.6
C. 5.9
告 無 =




⑦
D. 6.9 02 32g(ma
E. 4.3
=

晉
44 01g/mol E5
(02 = .



= 1 17.




ra


Chapter
翘 ]BEe ]]I11
1
筑 rr =
4 27
.




14. Based on molecular mass and dipole moment of the five compounds in the table below,
which should have the highest boiling point?




←
moment
I
>
- highest dipole

⑦
A. CH3CN
B. CH3CH2CH3
C. CH3OCH3
D. CH3Cl
E. CH3CHO
15. Of the following substances, only has London dispersion forces as its only
intermolecular force.
A. CH3OH
B. NH3
C. H2S
⑨ D. CH4
E. HCl
16. Which one of the following should have the lowest boiling point?
ㅐ , ,
* ,
A. PH3 H H H
B. H2S ㅣ
… " ↑
C. HCl H
⑨ D. SiH4 H C -



N
サーレー C C GGC
- -

-
-




E. H2O ㅣ

↓ ↓ . ↓
"
∞ 17. In which of the following
ㅐ molecules is hydrogen bonding likely to be the most significant
component of the total intermolecular
H H H forces?
시 시 H ㅣ
H H
A. CH4 ㅣ
ㅣ
ㅣ




B. C5H11OH H C
-

C CLCE O H
- - - -




C. C6H13NH2 1 1 "

① D. CH3OH H H i ↓ H

⑨ E. CO2
18. Of the following substances,
ㅐ only has hydrogen bonding.
ㅣ

H c o H c
o
-




0
- -

-
-




\

ㅐ

Información del documento

Subido en
6 de octubre de 2026
Número de páginas
19
Escrito en
2026/2027
Tipo
Examen
Contiene
Preguntas y respuestas
$36.49

¿Documento equivocado? Cámbialo gratis Dentro de los 14 días posteriores a la compra y antes de descargarlo, puedes elegir otro documento. Puedes gastar el importe de nuevo.
Escrito por estudiantes que aprobaron
Inmediatamente disponible después del pago
Leer en línea o como PDF

Seller avatar
Los indicadores de reputación están sujetos a la cantidad de artículos vendidos por una tarifa y las reseñas que ha recibido por esos documentos. Hay tres niveles: Bronce, Plata y Oro. Cuanto mayor reputación, más podrás confiar en la calidad del trabajo del vendedor.
VEVA2K
4.1
(692)
Vendido
2866
Seguidores
2415
Artículos
4515
Última venta
2 semanas hace



Por qué los estudiantes eligen Stuvia

Creado por compañeros estudiantes, verificado por reseñas

Calidad en la que puedes confiar: escrito por estudiantes que aprobaron y evaluado por otros que han usado estos resúmenes.

¿No estás satisfecho? Elige otro documento

¡No te preocupes! Puedes elegir directamente otro documento que se ajuste mejor a lo que buscas.

Paga como quieras, empieza a estudiar al instante

Sin suscripción, sin compromisos. Paga como estés acostumbrado con tarjeta de crédito y descarga tu documento PDF inmediatamente.

Student with book image

“Comprado, descargado y aprobado. Así de fácil puede ser.”

Alisha Student

Preguntas frecuentes