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Scilla [17]
4 years ago
7

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

um equation.
CO2(g) <----> CO2(aq) The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. K=0.032 M/atm
For a CO2 partial pressure of 4.4×10-4 atm in the atmosphere, what is the pH of water in equilibrium with the atmosphere?
Note: Since carbonic acid is primarily dissolved CO2, the concentration of H2CO3 can be taken as equal to that of dissolved CO2.
My online homework is due tonight and I am hopelessly lost on how to solve this problem. :(
Chemistry
1 answer:
Dvinal [7]4 years ago
6 0
C = pK 
<span>C = 4.4E-4*0.032 = about 1.41E-5 </span>

<span>H2CO3 ==> H^+ + HCO3^- </span>

<span>k1 = (H^+)(HCO3^-)/(H2CO3) </span>
<span>(H^+)= (HCO3^-) = x </span>
<span>(H2CO3) = 1.41E-5 </span>
<span>Solve for x = (H^+) and convert to pH. </span>
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3 years ago
A tank contains an ideal gas mixture of 5 g of O2 and 8 g of CO2 at 160kPa and specified temperature. If O2 were separated from
erica [24]

Answer:

74 or 74 kPa.

Explanation:

Hello,

In this case, based on the initial information, it is seen that the oxygen and the carbon dioxide form the mixture at 160 kPa, thus, by isolating the oxygen, its pressure will be equal to its initial partial pressure because it gets isolated, hence, we compute its molar fraction as:

x_{O_2}=\frac{5gO_2*\frac{1molO_2}{32gO_2} }{5gO_2*\frac{1molO_2}{32gO_2} +8gCO_2*\frac{1molCO_2}{44gCO_2} } =0.46

Therefore, its initial pressure turns out:

p_{O_2}=160kPa*0.46=73.9kPa

Such pressure will be the oxygen's pressure once it is isolated. Finally, considering the request, the answer will be just 74 (by rounding to the nearest integer and without units).

Best regards.

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ioda

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16

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