In which solution(s) will a precipitate form? i. 0.00090 g of na2cro4 (molar mass = 162.0 g/mol) is added to 225 ml of 0.00025 m agno3. ksp ag2cro4 =1.1 x 10-12 ii. 0.100l of 0.0015 m mgcl2 and 0.200l of 0.012 m naf. ksp mgf2 = 3.7 x 10-8

I. Solution will form a precipitate.

II. Solution will not form a precipitate.

Explanation:

I. 0,00090g of Na₂CrO₄ are:

0,00090g×= 5,56x10⁻⁶ moles of Na₂CrO₄≡ moles of CrO₄²⁻.

As volume of solution is 225mL:

5,56x10⁻⁶ moles of CrO₄²⁻ / 0,225L = 2,47x10⁻⁵ M of CrO₄²⁻.

The molar concentration of Ag⁺ is 0,00025M.

Ag⁺ with CrO₄²⁻ ions produce:

Ag₂CrO₄(s) ⇄ 2Ag⁺(aq) + CrO₄²⁻(aq)

Where kps is:

kps = 1,1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Replacing:

1,1x10⁻¹² = [0,00025M]² [2,47x10⁻⁵M]

1,1x10⁻¹² = 1,5x10⁻¹²

As the product of [Ag⁺]² [CrO₄²⁻] is higher than kps, the ions will decrease it concentration producing the precipitate.

II. Total volume of the solution is 0,300L, that means concentration of MgCl₂ and NaF will be:

0,0015M MgCl₂×= 0,0005M MgCl₂≡ Mg²⁺

0,012M NaF×= 0,008M NaF ≡ F⁻

F⁻ and Mg²⁺ ions produce:

MgF₂(s) ⇄ 2F⁻(aq) + Mg²⁺(aq)

Where kps is defined as:

kps = 3,7x10⁻⁸ = [F⁻]² [Mg²⁺]

Replacing:

3,7x10⁻⁸ = [0,008M]² [0,0005M]

3,7x10⁻⁸ = 3,2x10⁻⁸

As kps is higher than [F⁻]² [Mg²⁺], the ions will not form the precipitate.

I hope it helps!

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml