Because that is the outer ring.
After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml
<h3>Subtraction of Numbers</h3>
Given Data
- Volume of Hexane dispensed = 40ml
Let us compute the amount of excess hexane/ the volume that will remain
Remainder = The difference in volume dispensed and the volume needed
Remainder = 40-5
Remainder = 35 ml
The remainder is 35ml
Learn more about subtraction of numbers here:
brainly.com/question/4721701
Answer:
The answer is "2%"
Explanation:
Equation:


Formula:
![Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BNO_2%5E%7B-%7D%5D%7D%7B%5BHNO_2%5D%7D)
Let
at equilibrium

therefore,
![[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%202.0%5Ctimes%2010%5E%7B-2%7D%20%5C%20M%20%3D%200.02%20%5C%20M)
Calculating the % ionization:
![= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%5BH%5E%7B%2B%7D%5D%7D%7B%5BHNO_2%5D%29%7D%20%5Ctimes%20100%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.02%7D%7B1%7D%5Ctimes%20100%20%5C%5C%5C%5C%3D%202%5C%25%5C%5C%5C%5C)
The volume of H₂O = 5 L
<h3>Further explanation</h3>
Given
5L of H₂ and 3L O₂
Reaction
2H₂ (g) + O₂(g) ⇒2H₂O(g)
Required
The volume of H₂O
Solution
Avogadro's hypothesis:
<em>In the same T,P and V, the gas contains the same number of molecules </em>
So the ratio of gas volume will be equal to the ratio of gas moles
mol H₂ = 5, mol O₂ = 3
From equation, mol ratio H₂ : O₂ = 2 : 1, so :

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L
Answer:
C overgrazing is the answer