<h3>
Answer:</h3>
2.809 L of H₂SO₄
<h3>
Explanation:</h3>
Concept tested: Moles and Molarity
In this case we are give;
Mass of solid sodium hydroxide as 13.20 g
Molarity of H₂SO₄ as 0.235 M
We are required to determine the volume of H₂SO₄ required
<h3>First: We need to write the balanced equation for the reaction.</h3>
- The reaction between NaOH and H₂SO₄ is a neutralization reaction.
- The balanced equation for the reaction is;
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
<h3>Second: We calculate the umber of moles of NaOH used </h3>
- Number of moles = Mass ÷ Molar mass
- Molar mass of NaOH is 40.0 g/mol
Moles of NaOH = 13.20 g ÷ 40.0 g/mol
= 0.33 moles
<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
- From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
- Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
- Thus, Moles of H₂SO₄ = moles of NaOH × 2
= 0.33 moles × 2
= 0.66 moles of H₂SO₄
<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
- When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
- That is; Volume = Number of moles ÷ Molarity
In this case;
Volume of the acid = 0.66 moles ÷ 0.235 M
= 2.809 L
Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.
Atomic mass Hg = 200.59 u.m.a
200.59 g --------------- 6.02x10²³ atoms
( mass Hg ) ----------- 1.20 x10²² atoms
mass Hg = ( 1.20x10²² ) x 200.59 / 6.02x10²³
mass Hg = 2.407x10²⁴ / 6.02x10²³
= 3.998 g of Hg
hope this helps!
Two German chemists, Justus von Liebig (1803–1873) and Friedrich Wöhler (1800–1882), were responsible for the emergence of organic chemistry in the early nineteenth century. Their quantitative analytical methods helped establish the constitution of newly isolated and synthesized carbon compounds.
The particles in a solid are tightly packed and locked in place. Although we cannot see it or feel it, the particles are vibrating in place.
As these molecules heat up, they will vibrate more vigorously, and will eventually turn to water, then gas.
