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guapka [62]
2 years ago
15

25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?

Chemistry
1 answer:
MAVERICK [17]2 years ago
7 0

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

#SPJ1

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Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

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q=[q_1+q_2]

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where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

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Now put all the given values in the above formula, we get:

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

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Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

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\Delta H = enthalpy change = ?

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\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

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