As the combustible materials burn, some of the chemical energy is transformed into heat energy, and some are transformed into light energy. Light energy, also known as radiation or electromagnetic energy, is a type of kinetic energy that takes the form of visible light waves, such as the light from a match.
Do all substances dissolve in water? Kids explore the varying levels of solubility of common household substances in this fun-filled experiment!
Materials Needed:
4 clear, glass jars filled with plain tap water
Flour
Salt
Talcum or baby powder
Granulated sugar
Stirrer
Step 1: Help your child form a big question before starting the experiment.
Step 2: Make a hypothesis for each substance. Perhaps the salt will dissolve because your child has watched you dissolve salt or sugar in water when cooking. Maybe the baby powder will not dissolve because of its powdery texture. Help your child write down his or her predictions.
Step 3: Scoop a teaspoon of each substance in the jars, only adding one substance per jar. Stir it up!
Step 4: Observe whether or not each substance dissolves and record the findings!
Your child will likely note that that sugar and salt dissolve, while the flour will partially dissolve, and the baby powder will remain intact. The grainy crystals of the sugar and salt are easily dissolved in water, but the dry, powdery substances are likely to clump up or remain at the bottom of the jar.
As you can see, the scientific method is easy to work into your child’s scientific experiments. Not only does it increase your child’s scientific learning and critical thinking skills, but it sparks curiosity and motivates kids as they learn to ask questions and prove their ideas! Get started today with the above ideas, and bring the scientific method home to your child during your next exciting science experiment
Answer:
The reaction type is double displacement
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
