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GalinKa [24]
3 years ago
13

Show your work including units in your answer. Calculate the molar mass of Li4OH

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
4 0
Empirical formula: Li4OH
Answer:
1 mole: 44.771 g
1 gram = 0.022 mole

Explanation:

Element: Li
Percentage by mass: 62.01%
Number of atoms: 4
Mass of atom;: 6.941

Element: O
Percentage by mass: 35.74%
Number of atoms: 1
Mass of atom: 15.9994

Element: H
Percentage by mass: 2.25%
Number of atoms: 1
Mass of atom: 1.00794

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marishachu [46]

protons and electrons are both always the atomic number which is 9 in this case.

For neutrons you subtract the atomic number (9) from the weight of the atom (18.998) some teachers will want you to round to the nearest whole (19). We do this because the number of protons is the atomic number so if you subtract the protons from the whole weight of the atom you would have the electrons and neutrons left. Since electrons weigh so little we don't have to subtract them. Weighing neutrons and electrons would be like weighing an elephant (neutrons) and then putting one marshmallow on the scale (electron).  

5 0
3 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
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I can't come up with a good explanation.

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