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Alik [6]
3 years ago
11

The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is

Chemistry
1 answer:
Feliz [49]3 years ago
5 0
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
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Molecular equation

K2SO4(aq) + Ba(NO3)2(aq)  → BaSO4(s) + 2 KNO3(aq)?

This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:

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In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.

The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.

Learn more about precipitation here : brainly.com/question/1783904

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