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Jlenok [28]
3 years ago
11

Question 7 (1 point)

Chemistry
1 answer:
Anastasy [175]3 years ago
5 0

Answer:

2.28bar

Explanation: Boyle's law P1V1=P2V2 manipulate formula in favor of V2 the new formula should beP1V1/V2

0.895*318/125=2.2768 but to same sigfig it is 2.28

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Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na
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Answer:

pH=13

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

HA\rightleftharpoons H^+ + A^-

Which occurs in aqueous phase, therefore, the law of mass action is written by:

Ka=\frac{[H^+][A^-]}{[HA]}

That in terms of the change x due to the reaction's extent we can write:

1x10^{-20}=\frac{x*x}{0.1M-x}

But we prefer to compute the Kb due to its exceptional weakness:

Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}}  =1x10^{-6}

Next, the acid dissociation in the presence of the base we have:

Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}

Whose solution is x=0.0999M which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

pOH=-log([OH^-])=-log(0.0999)=1

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13

Regards.

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