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:
![[Pb^{2+}]=3.9 \times 10^{-2}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D3.9%20%5Ctimes%2010%5E%7B-2%7DM)
this is the concentration required to initiate precipitation
Explanation:
⇄
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
![Ip=[Pb^{2}][2Cl^-]^2=Ksp](https://tex.z-dn.net/?f=Ip%3D%5BPb%5E%7B2%7D%5D%5B2Cl%5E-%5D%5E2%3DKsp)

lets solubility=S
![[Pb^{2+}] = S](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%20%3D%20S)
![[Cl^-]=2S](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2S)
![Ksp=[Pb^{2+}]\times [Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%5BCl%5E-%5D%5E2)


![S=\sqrt[3]{\frac{Ksp}{4} }](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BKsp%7D%7B4%7D%20%7D)

this is the concentration required to initiate precipitation
Answer:
the catalyst is the two gray dots
Answer:
It's
Explanation:
Pbo + c
= pb + co2
While balancing it becomes
2pbo + c = 2pb + co2
(I was also doing same qn)