Answer:
<em>The pH of the solution is 7.8</em>
Explanation:
The concentration of the solution is 0.001M and the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:
[A-] = 0.001M - 0.0002M = 0.0008M
Also, the Henderson-Hasselbalch equation is
![pH = pKa + Log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20Log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation
Considering the Charles's law, the sample of carbon dioxide gas will occupy 308.72 mL.
<h3>Charles's law</h3>
Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.
Mathematically, Charles's law states that the ratio between volume and temperature will always have the same value:
Considering an initial state 1 and a final state 2, it is fulfilled:
<h3>Final volume in this case</h3>
In this case, you know:
- V1= 250 mL
- T1= 25 C= 298 K (being 0 C=273 K)
- V2= ?
- T2= 95 C= 368 K
Replacing in Charles's law:
Solving:
<u><em>V2= 308.72 mL</em></u>
Finally, the sample of carbon dioxide gas will occupy 308.72 mL.
Learn more about Charles's law:
brainly.com/question/4147359
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Out of the choices given, the part of the atom that forms chemical bonds are the outermost electrons. The correct answer is B.
Work out the number of moles in
100.00 grams of the oxide.
For nitrogen: The atomic mass of N is 14.0067, and we have 36.84 g N:
36.84 g N14.0067 g N/mol N=2.630 mol N
For oxygen: The atomic mass of O is
15.9994, and we have
100.00−36.84=63.16 g O:
63.16 g N 15.9994 g N/mol N=3.948 mol N
Now the ratio 3.958 2.630 is very close to
1.5=32
. So we conclude that the gas has three moles
O to two moles N making the empirical formula
N2O3.
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