門 ·0225mLx.1M = 2,25 × 10-3 mol NaOH 1.125 x 103 mul Acid A 1125×15 IL A Q 68 D D * AS AS SUFF ASUN! + -01125M + yel + 129 molar any) = mots Acit st ·0/125201 = (.129 Nolar Nars 16 F + > 17-1.44 4 १५.०५५ 1.125×103 = -3 125 molar ma Molar Mass = 106 g/md pure, ubstances mixed AST, V,i- A 4x² I 2 ,ok o D 4(096) ~ksp C +25 2, (ok) 2n + гон x (2x+) = 111 I (ol both) (0/60+23] 4080-11 0 X 00101 25 (0080+5)/45) = 3.00 × 10 17 (0.032,4,2) 4.0725 5.2851×150 5.78645×105 2-1773210-4 2.732x157 (.01901(42) (.0020) (4)
門 ·0225mLx.1M = 2,25 × 10-3 mol NaOH 1.125 x 103 mul Acid A 1125×15 IL A Q 68 D D * AS AS SUFF ASUN! + -01125M + yel + 129 molar any) = mots Acit st ·0/125201 = (.129 Nolar Nars 16 F + > 17-1.44 4 १५.०५५ 1.125×103 = -3 125 molar ma Molar Mass = 106 g/md pure, ubstances mixed AST, V,i- A 4x² I 2 ,ok o D 4(096) ~ksp C +25 2, (ok) 2n + гон x (2x+) = 111 I (ol both) (0/60+23] 4080-11 0 X 00101 25 (0080+5)/45) = 3.00 × 10 17 (0.032,4,2) 4.0725 5.2851×150 5.78645×105 2-1773210-4 2.732x157 (.01901(42) (.0020) (4)
Chapter24: Introduction To Spectrochemical Methods
Section: Chapter Questions
Problem 24.18QAP
Related questions
Question
Hello I need help clarifying how this person got a molar mass 106 g/mol. The image shows the how person got the answer, I am just still confused.
![門
·0225mLx.1M = 2,25 × 10-3 mol NaOH
1.125 x 103 mul Acid
A
1125×15
IL
A
Q
68
D
D
*
AS AS SUFF ASUN!
+
-01125M
+
yel
+
129
molar any) = mots Acit
st
·0/125201 = (.129
Nolar Nars
16
F
+
>
17-1.44
4 १५.०५५
1.125×103 =
-3
125
molar ma
Molar Mass = 106 g/md
pure, ubstances mixed AST, V,i-
A
4x²
I 2
,ok o
D
4(096) ~ksp
C
+25
2, (ok) 2n +
гон
x (2x+) = 111
I
(ol both)
(0/60+23]
4080-11
0
X
00101
25
(0080+5)/45) = 3.00 × 10 17
(0.032,4,2)
4.0725
5.2851×150
5.78645×105
2-1773210-4
2.732x157
(.01901(42)
(.0020) (4)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c9398a2-c49f-444d-8855-e495f02ff958%2Fd136df0b-ede3-4de7-9adc-572408a801aa%2Fpmg3ys_processed.jpeg&w=3840&q=75)
Transcribed Image Text:門
·0225mLx.1M = 2,25 × 10-3 mol NaOH
1.125 x 103 mul Acid
A
1125×15
IL
A
Q
68
D
D
*
AS AS SUFF ASUN!
+
-01125M
+
yel
+
129
molar any) = mots Acit
st
·0/125201 = (.129
Nolar Nars
16
F
+
>
17-1.44
4 १५.०५५
1.125×103 =
-3
125
molar ma
Molar Mass = 106 g/md
pure, ubstances mixed AST, V,i-
A
4x²
I 2
,ok o
D
4(096) ~ksp
C
+25
2, (ok) 2n +
гон
x (2x+) = 111
I
(ol both)
(0/60+23]
4080-11
0
X
00101
25
(0080+5)/45) = 3.00 × 10 17
(0.032,4,2)
4.0725
5.2851×150
5.78645×105
2-1773210-4
2.732x157
(.01901(42)
(.0020) (4)
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