⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠
Text Elements
Atom
Formula Unit
molecule
molecule
given
wanted
given units
1mol = 6.02 * 10^23 ( atoms, molecules, formula units )
15.0mol of Cu
Atoms
mol of Cu
6.022*10^23
1
9.033 * (10^24) atoms we only have 3 sig figs so 9.03
mol
3.8 * 10^14 molecules
molecule
6.022*10^23
1
6.31019595E-10 we have 2 sig figs so 6.3E-10 mol
32.00
12.01
2.02
40.08 + 79.90*2 40.08 + 159.8 199.88 g/mol = 1 mole of CaBr2
Sr(NO3)2 Sr(N2O6) 87.6 + 14.0 + 14.0 + (16.0*6) 211.6 g/mol = 1 mole of SR(NO3)2
1.25 mol
mol
1
gH2O
1.01+1.01+16.00=18.02
18.02
1.25*18.02 =22.5
mol
25g
g
22.99+ 35.5 = 58.44
58.44
1
(25*1)/58.44=.43 mol
mol
L
0.0580 mol
22.4
1
1.30
25.5g
1 mol
58.44g
1 mol
6.02 * 10^23 formula units
25.516.02E23/58.44/1 2.63 * 10^23
1 mol
12.0 + (35.5*4) = 154
154g
78g
6.02*10^23 molecules
1 mol
7816.02E23/154/1
3.04*10^23 molecules
Ca = 40.08 C = 12.01 O = 16.00 40.08 + 12.01 + (3 * 16.00) 100.09
40.08/100.09 - 0.400439604356 12.01/100.09 - 0.119992007194 48.00/100.09 - 0.4796 Ca - 40.04% C - 12.00% O - 47.96%
simplest
H2O2
HO
40.05g S 59.95g O
40.05g
mol
g
1
40.05*1/32.07 1.25
32.07
59.95gO
1 mol
16.00 gO
O
59.95*1/16.00 3.75
3.75/1.25 3
1.25/1.25 1
SO3
43.64g
1 mol
30.97g
56.36g
1 mol
16.00g
56.36*1/16.00 3.52
43.641/30.97 1.41/1.41=1 12=2
3.52/1.41=2.5 2.5*2 = 5
P2O5
actual
NO2
CH4N
12.01+(1.01*4)+14.01
30.06
C2H8N2
58.8g
1 mol
12.01g
9.8g
1 mol
1.01g
31.4g
1 mol
16g
4.90mol
9.70mol
1.96mol
1.96
1.96
1.96
2.5 5
4.95 10
1 2
EF:C5H10O2 5* 12.01 101.01 216 32+10.1+60.05=102.15 102.15*2=204 C10H20O4
coefficient
conversion factor
2mol H2
1 mol O2
1 mol O2
2 mol H2O
2 mol H2O
2 mol H2
4.2 mol H2
G
W
mol H2
mol O2
1
2
2.1 mol O2
W
G
.45mol AgNO3
mol AgNO3
mol AgCl
2
2
gAgCl
molAgCl
1
143.32
107.87+35.45=143.32
.45*143.32=64
G
W
gNaOH
molNaOH
14.5gNaOH
22.99+16+1.01=40
40.00
1
molNaOH
molBa(OH)2
1
2
gBa(OH)2
molBa(OH)2
137.38+(162)+(1.012)=171.4
171.4
1
14.511*171.4/40/2/1=31.1gBa(OH)2
4.20mol H2
1 mol H2O
2 mol H2
18.02g H2O
1 mol H2O
75.7g H2O this is the theoretical yield
75.7
60/75.7=.793 * 100 = 79.3%
4.57g Na(OH)2
1 mol
22.99+2(16+1.01)
40.00
mol NaOH
1 Ba(OH)2
2
1 mol BaOH2
g BaOH2
171.35
9.79g Ba(OH)2
8.50/9.79
86.8%
Embedded files
733bf7554226c1529539a6cdbc162894252bbedf: page=1 34f68049735cbebe1b438c403e15952624bb293f: page=2 3e7dcbf589b2da4aad141b1a5c2a181d5241f51b: page=3 9def0f91a3f35c6dda228f4767d5ef70c382ff33: page=4 168db1b582d40de7313e40caa13caf27fb468a75: page=5 67056c5abc4c99e8e3a3f3f7ec6473c83e04dca4: page=6 384b930883e0354ad245e042a36f7210c8d33e86: page=7 2ed7b6fb139bde664f8e12598b022c069cbd97f9: page=8 1178fcfe438e8adcb285f414f920e22dba569399: Pasted Image 20240408112255_776.png ea405d0d731c7669cb3e64573795dfda68e536a8: Pasted Image 20240408113111_801.png
