Answer:
P=atm

Explanation:
The problem give you the Van Der Waals equation:

First we are going to solve for P:


Then you should know all the units of each term of the equation, that is:







where atm=atmosphere, L=litters, K=kelvin
Now, you should replace the units in the equation for each value:

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

Then operate the fraction subtraction:
P=

And finally you can find the answer:
P=atm
Now solving for b:




Replacing units:

Multiplying and dividing units,(please see the second photo below), we have:



<span>The type of bond that a
Phosphorous pentachloride have is an Ionic Bonding. It is a form of chemical
bond that encompasses the electrostatic attraction between oppositely charged
ions which serves as the primary interaction happening in ionic compound. Phosphorus
has 5 valence electrons and Chlorine has 7 valence electrons. Phosphorus contributes
1 electron to each chlorine and all the 6 achieve 8 electrons in the outer
shell thus creating an ionic bond.</span>
The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.
<h3>What is the molarity of 5% NaOH?</h3>
The molarity of 5% NaOH is 1.32 M
25 cc of NaOH neutralized 30cc of H₂SO₄ solution.
Equation of reaction is given below:
- 2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O
Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M
- Normality = Molarity × moles of H⁺ ions per mole of acid
moles of H⁺ ions per mole of H₂SO₄ = 2
Normality of H₂SO₄ = 0.55 x 2 = 1.1 N
In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.
Learn more about normality at: brainly.com/question/22817773
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Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)