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
Answer:
Explanation:
molar volume at STP=22.4 L
given volume=50.0 L
number of moles=given volume/molar volume
number of moles=50.0/22.4
number of moles=2.2
1 mole of helium =6.023*10^23 atoms
2.2 moles of helium =6.023*10^23*2.2=1.3*10^24
therefore 50.0 L of helium contain 1.33*10^24 atoms
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Answer:
The correct answer is - Statement I is true, Statement-II is false.
Explanation:
