What volume of 0.2500 m naoh is required to neutralize 10.00 ml of 0.4000 m h2so4

volume is 32.00 mL

Explanation:

the balanced equation for the neutralisation reaction is as follows

2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

molar ratio of NaOH to H₂SO₄ is 2:1

number of H₂SO₄ moles reacted - 0.4000 mol/L x 10.00 x 10⁻³ L = 0.00400 mol

according to the molar ratio

number of NaOH moles reacted - 2 x number of H₂SO₄ moles reacted

number of NaOH moles reacted - 2 x 0.004000 mol = 0.008000 mol

concentration of NaOH is - 0.2500 mol/L

concentration = number of moles of solute / volume

rearrange the equation

volume = number of moles / concentration

volume = 0.008000 mol / 0.2500 mol/L

volume = 0.03200 L

therefore volume is 32.00 mL

is has three carbons, so the answer is propylamine.

meth-

eth-

prop-

but-

pent-

hex-

hept-

oct-

non-

explanation:

lol idk man i guess u gotta ask bill nye the science guy