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!

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)