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xz_007 [3.2K]
3 years ago
13

An aluminum sample has a mass of 80.1 g and a density of 2.70g/cm3 . According to the date to what number of significant figures

should the calculate volume of the aluminum sample be expressed?
Chemistry
2 answers:
chubhunter [2.5K]3 years ago
4 0
3 Sig figs because both measurements have three sig figs. In a lab setting, you would round to your least precise measurement. So its useless to use a meter stick and a high precision scale to take measurement because you have to use the amount of Sig figs for your least precise tool
DIA [1.3K]3 years ago
3 0
80.1 g* (1 cm^(3)/ 2.70 g)= 29.666666... cm^(3).

The calculated volume should have 3 significant figures, so the final answer is 2.97*10^(1) cm^(3)

Hope this helps~
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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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