1) Determine the grams of phosphate in 0.430 g potassium dihydrogen phosphate and show that the resulting solution when 0.430 g potassium dihydrogen phosphate is dissolved in 100 ml (100 g) water is 3000 ppm (3000 mg/L) in PO43–.

3000ppm of PO₄³⁻

Explanation:

First, we need to calculate the moles of potassium dihydrogen phosphate, KH₂PO₄. These moles are equal to moles of phosphate, PO₄³⁻:

Moles KH₂PO₄ -Molar mass: 136.086g/mol-:

0.430g KH₂PO₄ * (1mol / 136.086g) = 3.16x10⁻³moles KH₂PO₄ = Moles PO₄³⁻.

Now, mass of PO₄³⁻ is:

Mass PO₄³⁻ -molar mass: 94.97g/mol:

3.16x10⁻³moles PO₄³⁻ * (94.97g / 1mol) = 0.300g of PO₄³⁻ = 300mg

In 100mL = 0.100L

300mg PO₄³⁻ / 0.100L =

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i