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.
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I think it’s The second option sorry if I’m wrong
Answer: it’s number 2 add a catalyst
Hopefully this helped :)
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
