The concentration of [H3O⁺]=2.86 x 10⁻⁶ M
<h3>Further explanation</h3>
In general, the weak acid ionization reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
Ka's value
![\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKa%20%5C%3A%20%3D%20%5C%3A%20%5Cfrac%20%7B%5BH%20%5E%20%2B%5D%20%5BA%20%5E%20-%5D%7D%20%7B%5BHA%5D%7D%7D%7D%7D)
Reaction
HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵
![\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BC_2H_3O%5E%7B2-%7D%5BH_3O%5E%2B%5D%5D%7D%7B%5BHC_2H_3O_2%5D%7D%7D%5C%5C%5C%5C1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B0.22%5Ctimes%20%5BH_3O%5E%2B%5D%7D%7B0.035%7D)
[H₃O⁺]=2.86 x 10⁻⁶ M
%yield = 88.5%
<h3>Further explanation</h3>
Given
Reaction
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Required
The percent yield
Solution
mol AgNO₃(MW=169,87 g/mol) :
= mass : MW
= 127 : 169.87
= 0.748
mol Ag from equation :
= 2/2 x mol AgNO₃
= 2/2 x 0.748
= 0.748
Mass Ag (theoretical) :
= mol x Ar Ag
= 0.748 x 108
= 80.784
% yield = (actual/theoretical) x 100%
%yield = 71.5/80.784 x 100%
<em>%yield = 88.5%</em>
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Answer:
167.980.
Explanation:
Hello!
In this case, since additions involving the display of the result with the appropriate number of significant figures involve the solution of the system without looking at them:

Considering that 156.325 is significant to the thousandths and 11.65498 to the hundred thousandths, we infer that the result should be taken to the thousandths as well as 156.325; thus, we obtain:

Because the second decimal nine is rounded to ten and therefore the 7 is taken to 8.
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