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Afina-wow [57]
3 years ago
8

How are igneous rocks formed?

Chemistry
1 answer:
mariarad [96]3 years ago
5 0
B) cooling And solidification of magma
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Explain the relationship between the mole, particles, grams, and Liters.
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Answer:A mole is an arbitrary number of molecules in a single unit - refer to avogadro's number. Essentially, 1 mole is 6.022x10^23 molecules for ALL molecules or atoms, however one must remember that not all atoms/molecules are the same size, this is where mass comes into play. When you measure out 2 grams of carbon powder, there will be a lot more molecules present than if you weighed out 2 grams of thorium powder; this is because carbon is much smaller - kind of like a car filled with clowns, one given car can hold a lot of small clowns but only a few big ones; so the same volume is occupied but the amount of substance (clowns) varies on their own size. The arbitrary mass (relative to the hydrogen atom) for a molecule is the sum of its atomic components' atomic masses; e. g. C2H6's will have 2x12.00 (carbon) + 6x1.01 (hydrogen) = ~30 grams / mole.

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8 0
2 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 
5 0
3 years ago
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