Calculate the volume of 0.0500 M H2SO4 solution required to completely react with 20.0 mL of 1.00 x 10-3 M NaOH.

0.2 mL

Explanation:

When an acid and a base react, they converge in a neutralization reaction. So the formed products are water and a salt. The equation is:

H₂SO₄ (aq) + 2 NaOH (aq) → Na₂SO₄(aq) + 2H₂O(l)

1 mol of sulfuric needs 2 moles of hydroxide to react.

Let's find out the moles of base:

Molarity = mol/ volume (L) → molarity . volume(L) = moles

We make the conversion  → 20 mL . 1L /1000mL = 0.020L

0.001 M . 0.020L = 2×10⁻⁵ moles

So the rule of three is: 2 moles of base react with 1 mol of sulfuric

Therefore 2×10⁻⁵ moles of base will react with (2×10⁻⁵  . 1) / 2 = 1×10⁻⁵ moles of acid. Let's find out the volume:

Molarity = mol /volume(L) → mol / Molarity = Volume(L)

1×10⁻⁵ mol / 0.05 M = 2×10⁻⁴ L.

If we want the value in mL, we make the conversion:

2×10⁻⁴ L . 1000 mL /1L = 0.2 mL

passing a magnet thought the

mixture , because iron gets attracted to magnet .

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"