<span>Suppose you plant a seed and observe that a tree of large mass grows from it. The tree achieves a final mass that changes very little for years afterward.<span>
</span>Answer: Of the options presented above the one that is true about the tree is answer choice B) Both anabolic and catabolic reactions took place in the seed and tree when it was young and growing, and both continue now even though the tree reached a stable mass.
I hope it helps, Regards.</span>
The overall attraction of this process is called hydrogen bonding in water which have very powerful bonds. These powerful bonds are known as covalent bonds and are formed when electrons are shared by atoms.
The sharing of these electrons occur when hydrogen atoms share an electron + an oxygen atom. I hope I was able to satisfyingly answer your question. If you have any more questionings based on the information, let me know! :)
Answer:
1) The correct step in the scientific method that Victor did is Construct a hypothesis.
2) Given mass and density, volume is calculated as mass divided by density.
Explanation:
1) Before doing the assay and make a graph with the results obtained, Victor should think what he wants to prove, so he should make a hypoythesis to test with the assay.
2) The formula of density is
density = mass/volume ⇒ density x volume = mass ⇒ volume = mass/density.
Answer:
-1.42, -0.375, 32.5% (.325), 3/8 (.375), √4 (2.0), 3 (3.0), 2³ (8.0)
Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
<span>b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (</span>dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)<span>
d. 1.0 m CaCl2 (</span>dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))<span>
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))
The largest number of particles has </span>1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.
