Avolume of 90.0 ml of a 0.550 m hno3 solution is titrated with 0.890 m koh. calculate the volume of koh required to reach the equivalence point.

55.61 mL

Explanation:

Given;

Volume of HNO₃ solution, V₁ = 90.0 mL

Molarity of the solution, M₁ = 0.550 M

Molarity of KOH, M₂ = 0.890 M

the reaction between the two is given as:

HNO₃ + KOH ⇒ KNO₃ + H₂O

Here,

1 mol of HNO₃ reacts with 1 mol of KOH

thus, we have

M₁V₁ = M₂V₂

here,

V₂ is the volume of KOH

on substituting the values, we get

0.550 × 90.0 = 0.890 × V₂

or

V₂ = 55.61 mL

Hence,

The volume of KOH required to reach the equivalence point = 55.61 mL

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b