Unit 1B, Test Review]
siew
1(9.0 % )+2(12.0% 4)+6(1.0%..)+7(16.0g/mm)=151.0 /ad
b) % (=24033x100% = 15,9%
c)3,21 ghydrate, I na kredrate !And Belion 103,0gBEGOU
- 151.0 shydrats" Ino londrests Iand Belson
= 2.19 g Belaou
e
@ 1939 Mag Poux 64.
0 0 =75.390
164,0 g Najpoy :
adecuasto..$.23x1824
CO2 I mol COL . . .
6 382 gCO2 In con 62
o 118g Nassosx(x22 22 molecules con
642 Ss mas 6.022102 atores
2 mas lalo ators
- 138 .2
= 8,98 x1023 atomes
& mass C in iron corpsunda moss a in Asce
:.: =304,8 g Asu 35.5 a 75,5 g
148.4 g Agu
mass Fe =134.8 g total-755gC2= 59.3 g fe
mol - 1
. . . 1.06ms
1.06 mol
- 35.5 sce
.59,3 gFe-Im r=1.06mol Fé .1.06mlEL
75,5 g x lunch or 2.13mince 2.13 moich:
[Fechal
mass.Ba in original compound= mass Ba in Balroy
R = 2.012 ,BaCroup 137.3g Beaca = 1,091, Ba
:b) mass o = 1.345 gtotal-1,011g Ba = .254 , o
c) 1,091, Bax ImeBar = . 007946mlBa .007845 mol Ba
007946 mon .
254 g x me 01588 malo
887946 m .
= 01588 mano...
(BaO)
a) .843 Langairx l.18g dueano 1995 g dire air.
by mass of empty flask =187,20 97.995 g = 156.71
c) mass of unknown gas=158,08g-156.71g=1,37g
di moler mass=1:32 =40,0 %
e) % enor = 140x21-440%/ _=8,9%
44.0 %
- mas 392 mol
in iso l . Ima py
124,0gPy 2mil (as(poudz 310.3 g CagCroudz= 75.1
1mol Pu Iman caglpoud Cazlloudz.
b)1gSiol. ImoSion 10 moico 26.00 = 940.g(0
ul ca - 28.0
:() % . 75.6 g actual
Yield = x100% = 80.4 %
94.0 theorefree!
. . . 601a SiO2 6 molsio Imapy .
.:d) massPy=45.6gSiO2,ImalS:Ur Imolhy 1240spy=15.7gfy
• Massby= 38.2g Cx lm
ncilmarc ilmalPy 124,0 gPy =39.4.gpy
n o
Since the SiO2 formsless. Py, Sioris the limiting
: reactant and a maximum of 15.1, Pe can form,
g. 10mooc mul py ..
SS Co. Ima (oz Imoi CaCO3 loo.islacos,se
- 44.0 gCO2 imalou Iwasralo,
8.99g Glozi
3 x100% =48,1%
3 in sample.- 18.7 g total
: % SiO2 = 100.% -48.1% =51.9 % .:
GOO a) mass H in anilinea mass H in product the
= 6,63g H2o,20 st = 737 gH :
mass N in aniline = mass N in product Ne ...
. . gN . . . . . . . . . .
.:28.0
. = 146 a Nox2 28.0 g N . !
A = 1.46 a N : . : .
mass (= 9.71 9 -1737 g-1.46g=7.51 ..
Mul C = 7.51
. . a C Imola
. 12.
0 6 .626 mol C = 6
- .626 malc. 626 mol c . .
104 more
mal H = 737. H. Imolt
.? lo g.ne = 737 mal H .
I
.
2 1737much .7 :
you mor 1 . CLAN6km
. 104 mol
=1:46 3 Nx 1 mol N = 104mln.
14.0
. . . s W .104 mol N. . .104 mar a
low
: :
Molar. mass .
empirical mass .
n 93 ama
93 ama
mass. O = 750. g Kozx 1molKom 3 moi Onx320 g02
. 71.1 gko, 4 maxko, Imol O2
X 253 g O
:* : The maximum amount of oi that could be produced is more
:: : than what hasbeen produced, so not all of the Koz hasreacted.
# additional O to be produced =253 g =195g = 58.g.02.
mass Ko z reacted=!95gOzxIima ozx mlKor 11.1sko2 578gKoį
320go2 3 mol or mailko.
: massKoż remaining= 750.,Kon:-578gkoz= 172 zKoz
(12 a) No. If the pores are toolange, less solid will be captured
.:. . by the filter paper,which will decrease the mass.
by
. : Yes. Without rousing, the moisture on the solid will contain
dissolved ions. When thewater is removed, these ionswill
reform solid which will increase the mass.
paper.
c) No.
...the If some solid never makes it to the filter
most recorded willbe lower