A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.

Explanation:

1 mole HClO = 74.44g

0.0941g = $\frac{0.0941}{74.44}$ = 0.00126 moles

Concentration = no. of moles/volume in L

Hence, Concentration of HClO = 0.00126/ 0.250L

= 0.005M.

C1V1 =C2V2

0.005 × 250 mL = 0.2 × V2

Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.

3 0

3 years ago

In which group of the Periodic Table do most of the elements exhibit both positive and negative oxidation states?

Shtirlitz [24]

C i think but you should pick it anyway

4 0

3 years ago

At 80 ∘C, Kc=1.87×10^−3 for the reaction PH3BCl3(s)⇌PH3(g)+BCl3(g)

daser333 [38]

Answer:

A) he equilibrium concentration of PH3 = 0.0432M

B) he equilibrium concentration of BCl3 = 0.0432M

C) what is the minimum mass of PH3BCl3(s) that must be added to the flask to achieve equilibrium = 1.69g

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

6 0

3 years ago