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
HCl(aq) + KOH(s) --> KCl(aq) + H2O(l)
Answer:
235/92U+10n→144/54Xe+90/38Sr+2/10n
Explanation:
- The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;
235/92U+10n→144/54Xe+90/38Sr+2/10n
- In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
Answer : The labs were unable to reproduce the pharmaceutical company’s data.
Explanation : Any scientific claim must have reproducible experimental data. In this case, when the pharmaceutical company has the claim of reducing the cancer growth cells by 35% then by using the same manufacturing procedure for the drug and lab should be able to get this result. But they failed to match up with the results which clearly indicates that the labs were not able to produce the same results and hence they concluded that the pharmaceutical company's claims were invalid.
<u>Answer:</u> C) be hypertonic to Tank B.
<u>Explanation: </u>
<u>
The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution.
</u>
Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.
- If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
- If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
- If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.
In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>
Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>
Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.
