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 =

both adhesion and cohesion are attractions.

they are different in that adhesion is the force of attraction between the molecules of two different substances, whereas cohesion is the force of attraction between molecules of the same substance.

possible examples include capillary action, in which the adhesion between water and glass causes the water to rise in a narrow tube against the force of gravity.

because of hydrogen bonding, water is extremely cohesive.

this causes molecules on the surface of water to be drawn inward, which is why drops of water form beads on smooth surfaces.

hope this .

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