A construction worker is transporting pieces of plastic in two stacks with touching pieces. In Stack 1, both pieces of plastic are the same size. In Stack 2, the bottom piece is the same size as those in Stack 1, and the top piece is larger. The diagram above shows the energy of the molecules in the pieces of plastic before they touched.After a while, which of the two top plastic pieces will be cooler, and why?

Answers

Answer 1
Answer:

Answer:

The larger top plastic piece will be cooler than the smaller top plastic piece, because the energy that transferred to it was spread out over more molecules.

Explanation: I did the test


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how does the presence of a strong electrolyte in solution affect the colligative properties of a solution when compared to the same number of moles of a nonelectrolyte solute?

Answers

Answer:

  • lowered vapor pressure
  • higher boiling point

Explanation:

The colligative properties of a given solution can be defined as the properties of that solution that are dependent on the concentration of the molecules or ions of the solute in the solution, and not on the type or identity of that solute. Examples include:

1. vapor pressure lowering

2. boiling point elevation

3. freezing point depression

4. Osmotic pressure

In this case, vapor pressure would be lowered because with an electrolyte introduced into a solution, the number of solute particles would be larger because the solute particles dissociate into ions, thereby competing with the solvent molecules at the surface of the solution, which in turn reduces the rate at which the solvent evaporates and condenses. Vapor pressure is lower compared to a solution with the same number of moles of  nonelectrolyte solute.

The higher the number of ions in the solution, the greater the colligative properties of the solution will be impacted.

Final answer:

The presence of a strong electrolyte in solution affects the colligative properties differently than a non-electrolyte solute. Strong electrolytes dissociate into ions, increasing the number of particles in solution. This affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.

Explanation:

The presence of a strong electrolyte in solution affects the colligative properties differently than the same number of moles of a non-electrolyte solute. This is because strong electrolytes dissociate into ions when dissolved in solution, while non-electrolytes do not. The dissociation of strong electrolytes increases the total number of particles in solution, which affects colligative properties such as vapor pressure, boiling point elevation, freezing point depression, and osmotic pressure.

For example, let's compare a solution of 1 mole of sodium chloride (NaCl) to a solution of 1 mole of sucrose (C12H22O11). The sodium chloride will dissociate into Na+ and Cl- ions, which means there are now 2 particles in solution (1 Na+ and 1 Cl-) instead of just 1 molecule of sucrose. This higher particle concentration will result in a greater depression of the freezing point and elevation of the boiling point compared to the sucrose solution.

In summary, the presence of a strong electrolyte increases the number of particles in solution, leading to greater deviations in colligative properties compared to the same number of moles of a non-electrolyte solute.

Learn more about Colligative properties here:

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A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is 4.70 4.70 , what is the pH of the buffer?

Answers

Answer: The pH of the buffer is 4.61

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log((3.25)/(4.00))\n\npH=4.61

Hence, the pH of the buffer is 4.61

Suppose a scientist made a claim that all spontaneous reactions are exothermic. Whic of the following would provide the strongest challenge to their claim? Suppose a scientist made a claim that all spontaneous reactions are exothermic. Which of the following would provide the strongest challenge to their claim? a. An exothermic reaction which is not spontaneous
b. An endothermic reaction that only proceeds when coupled to an exothermic reaction
c. An endothermic reaction that only proceeds when a catalytst is present
d. An endothermic reaction which is not spontaneous
e. All of the above

Answers

Answer: Option (c) is the correct answer.

Explanation:

It is given that the scientist is claiming that all the spontaneous reactions are exothermic in nature.

And, it is known that when a reaction is spontaneous in nature then \Delta G is negative.

Now, the relation between Gibb's free energy, enthalpy and entropy is as follows.

               \Delta G = \Delta H - T \Delta S

So, when a catalyst is present in a chemical reaction then we do not need to give large amount of heat from outside. And, because of this the enthalpy of reaction will not be highly positive.

Hence, the value of \Delta G will result in a negative value which means the reaction is spontaneous.

Thus, we can conclude that an endothermic reaction that only proceeds when a catalytst is present, would provide the strongest challenge to their claim.

What else is produced during the replacement reaction of silver nitrate and potassium sulfate? 2AgNO3 + K2SO4 Ag2SO4 + ________

Answers

Answer:

c

Explanation:

Answer:

c

Explanation:

Identify which location in the periodic table you would have the
largest atomic radii.

Answers

Answer:

left to right across a period when it decreases and when it increases top to bottom in a group,

hope i helped

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Answers

Ba(IO₃)₂(s) partially dissociates in water  as Ba²⁺(aq) and IO₃⁻(aq).
                Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial          
Change        -X                +X             2X
Equilibrium                        X              2X

           Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰  = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
              X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L