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Feliz [49]
2 years ago
11

Suppose an alkane had 30 hydrogen atoms. How many carbon atoms would it have?

Chemistry
2 answers:
vesna_86 [32]2 years ago
5 0
The answer would be C - 14 atoms.
liberstina [14]2 years ago
3 0

Answer: c.  14 atoms

Explanation:

Saturated hydrocarbons are defined as the hydrocarbons in which a single bond is present between carbon and carbon atoms. The general formula for these hydrocarbons is C_nH_{2n+2} and they are called as alkanes.

Given: An alkane contains 30 hydrogen atoms

Thus we can write:

2n+2=30

2n=28

n=14

Thus the number of carbon atoms will be n which is 14.

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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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