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Coffee is a solution containing caffeine, acids, alkaloids, water, phenols, and many other chemicals. The solvent in the coffee solution is (1) __________ and the rest of the chemicals dissolved in the solvent are called (2)_________.


3. What is the difference between a saturated solution and a supersaturated solution?


4. Is the following molecule, polar or nonpolar?

5. 12 eggs can be referred to as one dozen eggs.
______________ particles can be referred to as one mole of particles.

6.
The pH of a strong acid might be _____ while the pH of a strong base might be _____
The pH of pure water is _____.

7. What would be the pH of .002 moles of HNO3 dissolved in 2 L of water?

8. A base has a pH of 8.5. What is the concentration of OH- ions in the solution?

Respuesta :

Solutes dissolve in solvents to form a solution. A saturated solution contains just as must solute as it can normally hold.

The solvent in the coffee solution is water and the rest of the chemicals dissolved in the solvent are called solutes. A solution is formed when a solute is dissolved in a solvent.

A saturated solution contains just as much solute as it can normally hold at a particular temperature while a supersaturated solution contains more solute than it can normally hold at a particular temperature.

A polar molecule contains covalent bonds between atoms having an electronegativity difference above 0.5. Such molecules are polar as electrons of the bond are drawn closer to the atom that is more electronegative.

According to Avogadro's law; 6.02 × 10^23 particles is referred to as one mole of particles.

A strong acid has a pH that may range from 0 - 3. A strong base has a pH of around 10 - 14. Water is a neutral substance and has a pH of 7.

From the information provided;

Number of moles of acid = 0.002 moles

Volume of solution=  2 L

Concentration of solution = number of moles/volume = 0.002 moles/2L = 0.001 M

pH = - log[H^+]

pH = - log[0.001 ]

pH = 3

From;

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 8.5

pOH = 5.5

pOH = - log[OH^-]

[OH^-] = Antilog[-5.5]

[OH^-] = 3.2 × 10^-6 M

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