I think the answer is C I think that bc I added 3 and the 2 which made it H5 and than multipled the 3 and the 2 which made it C6 than Cl
15 mins
~Savannah
Jokes from Savannah~
A man goes to the doctor, concerned about his wife’s hearing. The doctor says, “Stand behind her and say something and tell me how close you are when she hears you.”
The man goes home, sees his wife in the kitchen, cutting carrots on the countertop. About 15 feet away he says, “Honey, what’s for dinner?” Nothing. He gets halfway to her and repeats the same question. Nothing. Very concerned, he gets right behind her and asks again “What’s for dinner?”
She turns around and says “For the THIRD time, beef stew!”
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Q: What do you get when you cross a vampire with a snowman?
A: Frostbite!
Answer:
Molecular compound and ionic compound
Explanation:
Molecular compounds include a prefix on the first and second elements and a suffix on the second element, whereas Ionic compounds do not have prefixes and only include a suffix on the second element.
Answer:
The more polar the liquid, the more likely that it is miscible with water. The polarity of a liquid does not affect its miscibility with water. The less polar the liquid, the more likely that it is miscible with water. The more polar the liquid, the less likely that it is miscible with water.
Explanation:
hope it helps you
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