It has more density when it sinks because the water pushes away thats why people float they are less dense
N=24/12
n=2
where n= no. of moles
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)

[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:

![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)

=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
Answer:
Light, sound, and waves in the ocean are common examples of waves. Sound and water waves are mechanical waves; meaning, they require a medium to travel through. The medium may be a solid, a liquid, or a gas, and the speed of the wave depends on the material properties of the medium through which it is traveling.
Explanation:
hope this helps
Answer:
The acid dissociation constant, _Ka__, is a quantitative measure of acid strength
Explanation: