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Black_prince [1.1K]
3 years ago
11

There are two open cans of soda on the table. One can was just taken from the refrigerator and the other was taken from the cupb

oard. Upon opening the cans, which one loses carbon dioxide more quickly and why? A.The can from the refrigerator. Because it is colder and gases are less soluble in colder temperatures. B.The can from the refrigerator, because it is colder and gases are more soluble in colder temperatures. C.The can from the cupboard, because it is warmer and gases are less soluble in warmer temperatures. D.Neither, because the solubility of gases is affected only by pressure.
Chemistry
2 answers:
kvv77 [185]3 years ago
7 0

The correct answer is

C- The can from the cupboard, because it is warmer and gases are less soluble in warmer temperatures

:)

nika2105 [10]3 years ago
5 0
I think the correct answer from the choices listed above is option C. The can <span>from the cupboard will lose carbon dioxide more quickly because it is warmer and gases are less soluble in warmer temperatures. </span> Solubility of gases is a strong function of temperature and as well as pressure.
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First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2  so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄      ⇄     H⁺   + HC₂O₄⁻
0.0356 M            0          0
0.0356 - x            x          x
Ka1 = \frac{[H^+][HC2O4^-]}{[H2C2O4]} = x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
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3 years ago
Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

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